🧠:[<think>Okay, so I need to find the value of sin2x given that sin(π/4 - x) = 3/5. Let me think about how to approach this. First, I remember that there are trigonometric identities that relate angles subtracted or added. Specifically, the sine of a difference of angles. The formula for sin(A - B) is sinA cosB - cosA sinB. Maybe I can use that identity here. Let me write that down:sin(π/4 - x) = sin(π/4)cosx - cos(π/4)sinx.Since sin(π/4) and cos(π/4) are both √2/2, substituting those values in:(√2/2)cosx - (√2/2)sinx = 3/5.So, combining the terms:(√2/2)(cosx - sinx) = 3/5.Hmm, maybe I can factor that √2/2 out. Let me solve for cosx - sinx. Multiply both sides by 2/√2:cosx - sinx = (3/5)*(2/√2) = 6/(5√2).Let me rationalize the denominator:6/(5√2) = (6√2)/(5*2) = (3√2)/5.So, cosx - sinx = 3√2/5.Now, I need to find sin2x. I recall that sin2x = 2sinxcosx. So if I can find sinx and cosx, or maybe the product sinxcosx, I can compute sin2x.Alternatively, perhaps I can square both sides of the equation cosx - sinx = 3√2/5. Then, expand the left side and relate it to sin2x.Let me try that. Squaring both sides:(cosx - sinx)^2 = (3√2/5)^2.Left side expands to cos²x - 2sinxcosx + sin²x.Right side is (9*2)/25 = 18/25.So, cos²x + sin²x - 2sinxcosx = 18/25.But I know that cos²x + sin²x = 1, so substituting that in:1 - 2sinxcosx = 18/25.Then, subtracting 1 from both sides:-2sinxcosx = 18/25 - 1 = 18/25 - 25/25 = -7/25.Multiply both sides by -1:2sinxcosx = 7/25.But 2sinxcosx is exactly sin2x. Therefore,sin2x = 7/25.Wait, that seems straightforward. Let me check my steps again to make sure I didn't make a mistake.Starting from sin(π/4 - x) = 3/5. Applied the sine difference identity correctly, substituted the known values of sin(π/4) and cos(π/4). Then solved for cosx - sinx, rationalized the denominator. Then squared both sides, expanded, used the Pythagorean identity, solved for sin2x. All steps seem correct.Let me verify with an alternative approach. Suppose I let θ = π/4 - x. Then sinθ = 3/5. Then θ = π/4 - x, so x = π/4 - θ. Then 2x = π/2 - 2θ. Then sin2x = sin(π/2 - 2θ) = cos2θ. So sin2x = cos2θ. Since sinθ = 3/5, cos2θ = 1 - 2sin²θ = 1 - 2*(9/25) = 1 - 18/25 = 7/25. Therefore, sin2x = 7/25. Same result. So that checks out.Therefore, the answer should be 7/25.Final AnswerThe value of (sin2x) is boxed{dfrac{7}{25}}.</think>Given (sinleft( frac{pi}{4} - x right) = frac{3}{5}), we need to find (sin2x).First, we use the sine difference identity:[sinleft( frac{pi}{4} - x right) = sinleft( frac{pi}{4} right)cos x - cosleft( frac{pi}{4} right)sin x]Since (sinleft( frac{pi}{4} right) = cosleft( frac{pi}{4} right) = frac{sqrt{2}}{2}), we substitute these values:[frac{sqrt{2}}{2} cos x - frac{sqrt{2}}{2} sin x = frac{3}{5}]Factoring out (frac{sqrt{2}}{2}):[frac{sqrt{2}}{2} (cos x - sin x) = frac{3}{5}]Solving for (cos x - sin x):[cos x - sin x = frac{3}{5} cdot frac{2}{sqrt{2}} = frac{6}{5sqrt{2}} = frac{3sqrt{2}}{5}]Next, we square both sides:[(cos x - sin x)^2 = left( frac{3sqrt{2}}{5} right)^2]Expanding the left side and simplifying the right side:[cos^2 x - 2 sin x cos x + sin^2 x = frac{18}{25}]Using the Pythagorean identity (cos^2 x + sin^2 x = 1):[1 - 2 sin x cos x = frac{18}{25}]Solving for (sin 2x):[1 - frac{18}{25} = 2 sin x cos x][frac{7}{25} = 2 sin x cos x]Thus,[sin 2x = frac{7}{25}]The value of (sin2x) is boxed{dfrac{7}{25}}.,<think>Alright, let's see. I need to find sin2x given that sin(π/4 - x) = 3/5. Hmm, okay. So, first, maybe I should recall some trigonometric identities that could help here. The problem gives me a sine of a difference, and I need to find sine of double an angle. Let me jot down what I know.First, the sine of a difference identity: sin(A - B) = sinA cosB - cosA sinB. So, applying that to sin(π/4 - x), that would be sin(π/4)cosx - cos(π/4)sinx. Since sin(π/4) and cos(π/4) are both √2/2, right? So substituting that in, we get:sin(π/4 - x) = (√2/2)cosx - (√2/2)sinx = √2/2 (cosx - sinx). And that equals 3/5. So:√2/2 (cosx - sinx) = 3/5.Hmm, okay. Maybe I can solve for (cosx - sinx) first. Let's multiply both sides by 2/√2 to get rid of the coefficient on the left. That would give:cosx - sinx = (3/5) * (2/√2) = 6/(5√2). Let's rationalize the denominator here. Multiply numerator and denominator by √2:6√2 / (5*2) = (6√2)/10 = (3√2)/5. So, cosx - sinx = 3√2/5.Alright, so we have cosx - sinx = 3√2/5. Maybe I can square both sides here to use the Pythagorean identity later. Let's try that.(cosx - sinx)^2 = (3√2/5)^2. Expanding the left side:cos²x - 2sinx cosx + sin²x = (9*2)/25 = 18/25.Now, cos²x + sin²x = 1, so substituting that in:1 - 2sinx cosx = 18/25.Hmm, okay. Then, subtract 1 from both sides:-2sinx cosx = 18/25 - 1 = 18/25 - 25/25 = -7/25.Multiply both sides by -1:2sinx cosx = 7/25.But wait, 2sinx cosx is exactly sin2x. So, sin2x = 7/25. That seems straightforward. Let me check my steps again to make sure I didn't make a mistake.Starting with sin(π/4 - x) = 3/5. Applied the sine difference identity correctly. Plugged in sin(π/4) and cos(π/4) as √2/2. Factored that out, then multiplied both sides by 2/√2. Did the algebra to get cosx - sinx = 3√2/5. Then squared both sides. Expanded left side to cos²x - 2sinx cosx + sin²x. Replaced cos²x + sin²x with 1. So, 1 - 2sinx cosx = 18/25. Solving for sin2x gives 7/25. That seems right.Alternatively, maybe I can approach this problem using another identity. Let's see. If I let θ = π/4 - x, then sinθ = 3/5. Then, x = π/4 - θ. So, sin2x = sin(2(π/4 - θ)) = sin(π/2 - 2θ). Then, sin(π/2 - 2θ) = cos2θ. So, sin2x = cos2θ. Then, since sinθ = 3/5, cos2θ = 1 - 2sin²θ. Which would be 1 - 2*(9/25) = 1 - 18/25 = 7/25. So, same result. That confirms it. So, sin2x = 7/25.Therefore, the answer should be 7/25. Let me check one more time. Let's compute cosx - sinx = 3√2/5. Squared is 9*2/25 = 18/25. Then, 1 - sin2x = 18/25, so sin2x = 1 - 18/25 = 7/25. Yep, that's correct. Okay, confident now. The answer is 7/25.Final Answerboxed{dfrac{7}{25}}</think>Given (sinleft( frac{pi}{4} - x right) = frac{3}{5}), we need to find (sin 2x).First, we use the sine difference identity:[sinleft( frac{pi}{4} - x right) = sinleft( frac{pi}{4} right) cos x - cosleft( frac{pi}{4} right) sin x]Since (sinleft( frac{pi}{4} right) = cosleft( frac{pi}{4} right) = frac{sqrt{2}}{2}), we substitute these values:[frac{sqrt{2}}{2} cos x - frac{sqrt{2}}{2} sin x = frac{3}{5}]Factor out (frac{sqrt{2}}{2}):[frac{sqrt{2}}{2} (cos x - sin x) = frac{3}{5}]Multiply both sides by (frac{2}{sqrt{2}}) to isolate (cos x - sin x):[cos x - sin x = frac{3}{5} cdot frac{2}{sqrt{2}} = frac{6}{5sqrt{2}} = frac{3sqrt{2}}{5}]Square both sides to use the Pythagorean identity:[(cos x - sin x)^2 = left( frac{3sqrt{2}}{5} right)^2][cos^2 x - 2 sin x cos x + sin^2 x = frac{18}{25}]Using (cos^2 x + sin^2 x = 1):[1 - 2 sin x cos x = frac{18}{25}]Solving for (2 sin x cos x) (which is (sin 2x)):[1 - frac{18}{25} = 2 sin x cos x][frac{7}{25} = sin 2x]Thus, the final answer is (boxed{dfrac{7}{25}}).]

🔑:Since sinleft( frac {pi}{4}-xright)= frac {3}{5}, then sin2x=cosleft( frac {pi}{2}-2xright)=1-2sin^{2}left( frac {pi}{4}-xright)= frac {7}{25}． Therefore, the answer is: boxed{frac {7}{25}}． This solution involves using the trigonometric identity and the double-angle formula for cosine, combined with the given information to calculate the value. This question primarily tests the application of trigonometric identities and the double-angle formula for cosine in simplifying and evaluating trigonometric expressions, and it is considered a basic problem.

🧠:[<think>Okay, so I need to solve this matrix equation: A * X = B, where A is a 3x3 matrix, X is a 3x1 column vector, and B is another 3x1 column vector. The matrix A is given as:left(begin{array}{rrr}3 & -1 & 0 -2 & 1 & 1 2 & -1 & 4end{array}right)and B is:left(begin{array}{r}5 0 15end{array}right)So, the goal is to find the vector X = [x, y, z]^T such that when we multiply matrix A by X, we get B. I remember that to solve for X, we can use the inverse of matrix A, provided that A is invertible. The formula would be X = A^{-1} * B. So first, I need to check if matrix A is invertible. To do that, I should compute the determinant of A. If the determinant is not zero, then the matrix is invertible.Alright, let's compute the determinant of A. The determinant of a 3x3 matrix can be calculated using the rule of Sarrus or the general formula. Let me recall the formula for the determinant of a 3x3 matrix:For a matrix:left(begin{array}{ccc}a & b & c d & e & f g & h & iend{array}right)the determinant is a(ei - fh) - b(di - fg) + c(dh - eg).Applying this to matrix A:a = 3, b = -1, c = 0d = -2, e = 1, f = 1g = 2, h = -1, i = 4So determinant = 3*(1*4 - 1*(-1)) - (-1)*(-2*4 - 1*2) + 0*(-2*(-1) - 1*2)Let me compute each term step by step.First term: 3*(1*4 - 1*(-1)) = 3*(4 + 1) = 3*5 = 15Second term: -(-1)*(-2*4 - 1*2) = 1*(-8 - 2) = 1*(-10) = -10Wait, hold on. Let me check that again. The formula is -b*(di - fg). So here, b is -1, so it's -(-1)*( (-2)*4 - 1*2 ). So that's +1*( -8 - 2 ) = 1*(-10) = -10Third term: +c*(dh - eg) = 0*(...) = 0So total determinant = 15 -10 +0 = 5Since the determinant is 5, which is not zero, matrix A is invertible. Great, so we can proceed to find the inverse of A and multiply it by B to get X.Alternatively, another method is to use Gaussian elimination to solve the system of equations. Maybe I can try both methods to cross-verify.First, let me write down the system of equations represented by A * X = B.The matrix equation is:3x - y + 0z = 5-2x + y + z = 02x - y + 4z = 15So, writing these as equations:1) 3x - y = 52) -2x + y + z = 03) 2x - y + 4z = 15Hmm, maybe using substitution or elimination would work here. Let me try solving these equations step by step.From equation 1: 3x - y = 5. Let's solve for y:y = 3x -5Now, substitute y into equations 2 and 3.Equation 2: -2x + (3x -5) + z = 0Simplify:-2x + 3x -5 + z = 0 => x -5 + z = 0 => x + z = 5 => z = 5 - xEquation 3: 2x - (3x -5) + 4z = 15Simplify:2x -3x +5 +4z =15 => -x +5 +4z =15 => -x +4z =10Now, from equation 2 substitution, z =5 - x. Substitute this into equation 3:-x +4*(5 -x) =10Compute:-x +20 -4x =10 => -5x +20 =10 => -5x= -10 => x=2Then, substitute x=2 back into z=5 -x: z=5 -2=3Then, substitute x=2 into y=3x -5: y=6 -5=1So, x=2, y=1, z=3. Let me check if these values satisfy all three equations.Equation 1: 3*2 -1 =6 -1=5 ✔️Equation 2: -2*2 +1 +3= -4 +1 +3=0 ✔️Equation 3: 2*2 -1 +4*3=4 -1 +12=15 ✔️Perfect, so the solution is X = [2, 1, 3]^T.Alternatively, using matrix inversion. Let's see if I can compute A inverse and then multiply by B to get the same result.First, computing the inverse of A. Since the determinant is 5, which we already found, the inverse is (1/det) * adjugate matrix.The adjugate matrix is the transpose of the cofactor matrix.Let me compute the cofactor matrix of A.The cofactor C_{ij} = (-1)^{i+j} * M_{ij}, where M_{ij} is the minor, the determinant of the submatrix obtained by removing row i and column j.Matrix A:Row 1: 3, -1, 0Row 2: -2, 1, 1Row 3: 2, -1, 4Compute each cofactor:C11: (+) determinant of submatrix when row1, col1 removed:left(begin{array}{rr}1 & 1 -1 & 4end{array}right)det = (1*4) - (1*(-1)) =4 +1=5C11=5C12: (-) determinant of submatrix row1, col2:left(begin{array}{rr}-2 & 1 2 & 4end{array}right)det=(-2*4)-(1*2)= -8 -2= -10C12= -(-10)=10 (Wait, hold on. The cofactor is (-1)^{1+2} * M12. So (-1)^3 * (-10)= -1*(-10)=10Wait, the minor M12 is the determinant of the submatrix, which is (-2*4 -1*2)= -8-2= -10. Then C12= (-1)^{1+2}*(-10)= (-1)^3*(-10)= -1*(-10)=10. Yes.C13: (+) determinant of submatrix row1, col3:left(begin{array}{rr}-2 & 1 2 & -1end{array}right)det=(-2*(-1)) - (1*2)=2 -2=0C13=0Now row 2:C21: (-) determinant of submatrix row2, col1:left(begin{array}{rr}-1 & 0 -1 & 4end{array}right)det=(-1*4) - (0*(-1))= -4 -0= -4C21= (-1)^{2+1}*(-4)= (-1)^3*(-4)= -(-4)=4Wait, no: cofactor C21 is (-1)^{2+1} * M21. M21 is the determinant of the submatrix which is (-1*4 -0*(-1))= -4. Then C21= (-1)^3*(-4)= -1*(-4)=4. Correct.C22: (+) determinant of submatrix row2, col2:left(begin{array}{rr}3 & 0 2 & 4end{array}right)det=3*4 -0*2=12C22= (-1)^{2+2}*12=1*12=12C23: (-) determinant of submatrix row2, col3:left(begin{array}{rr}3 & -1 2 & -1end{array}right)det=3*(-1) - (-1)*2= -3 +2= -1C23= (-1)^{2+3}*(-1)= (-1)^5*(-1)= -1*(-1)=1Row 3:C31: (+) determinant of submatrix row3, col1:left(begin{array}{rr}-1 & 0 1 & 1end{array}right)det=(-1*1)-(0*1)= -1 -0= -1C31= (-1)^{3+1}*(-1)=1*(-1)= -1C32: (-) determinant of submatrix row3, col2:left(begin{array}{rr}3 & 0 -2 & 1end{array}right)det=3*1 -0*(-2)=3 -0=3C32= (-1)^{3+2}*3= (-1)^5*3= -3C33: (+) determinant of submatrix row3, col3:left(begin{array}{rr}3 & -1 -2 & 1end{array}right)det=3*1 - (-1)*(-2)=3 -2=1C33= (-1)^{3+3}*1=1*1=1So the cofactor matrix is:Row 1: 5, 10, 0Row 2: 4, 12, 1Row 3: -1, -3, 1Now, the adjugate matrix is the transpose of the cofactor matrix. So we need to transpose the above matrix.Transposing:Original cofactor:First column of cofactor becomes first row of adjugate:Row 1 of adjugate: 5, 4, -1Row 2 of adjugate: 10, 12, -3Row 3 of adjugate: 0, 1, 1Therefore, adjugate(A) is:left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right)Then, inverse of A is (1/determinant)*adjugate(A) = (1/5)*adjugate(A):So,A^{-1} = (1/5)*left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right)Which gives:First row: 1, 4/5, -1/5Second row: 2, 12/5, -3/5Third row: 0, 1/5, 1/5Now, to compute X = A^{-1} * B, where B is [5; 0; 15]So let's perform the multiplication:First component:(1*5) + (4/5*0) + (-1/5*15) = 5 + 0 -3 = 2Second component:(2*5) + (12/5*0) + (-3/5*15) =10 +0 -9=1Third component:(0*5) + (1/5*0) + (1/5*15) =0 +0 +3=3So X = [2; 1; 3], which matches the solution we found earlier by substitution. Great, so both methods give the same result. Therefore, the solution to the matrix equation is x=2, y=1, z=3. So the vector X is:left(begin{array}{r}2 1 3end{array}right)I think that's thorough. Let me just verify once again with the original matrix multiplication.Compute A * X:First row: 3*2 + (-1)*1 +0*3 =6 -1 +0=5 ✔️Second row: -2*2 +1*1 +1*3= -4 +1 +3=0 ✔️Third row:2*2 + (-1)*1 +4*3=4 -1 +12=15 ✔️Yep, everything checks out. So confident the answer is correct.Final AnswerThe solution to the matrix equation is boxed{left(begin{array}{r} 2 1 3 end{array}right)}.</think>To solve the matrix equation left(begin{array}{rrr}3 & -1 & 0 -2 & 1 & 1 2 & -1 & 4end{array}right) X = left(begin{array}{r}5 0 15end{array}right),we first check if the matrix ( A ) is invertible by computing its determinant. The determinant of ( A ) is calculated as follows:text{det}(A) = 3 left(1 cdot 4 - 1 cdot (-1)right) - (-1) left(-2 cdot 4 - 1 cdot 2right) + 0 left(-2 cdot (-1) - 1 cdot 2right) = 3 cdot 5 - (-1) cdot (-10) + 0 = 15 - 10 + 0 = 5.Since the determinant is non-zero, ( A ) is invertible. We then solve the system of equations using substitution:1. From ( 3x - y = 5 ), we get ( y = 3x - 5 ).2. Substituting ( y = 3x - 5 ) into the second equation ( -2x + y + z = 0 ), we get ( x + z = 5 ), so ( z = 5 - x ).3. Substituting ( y = 3x - 5 ) and ( z = 5 - x ) into the third equation ( 2x - y + 4z = 15 ), we solve for ( x ): -x + 4(5 - x) = 10 implies -5x + 20 = 10 implies x = 2. Substituting ( x = 2 ) back, we get ( y = 1 ) and ( z = 3 ).To verify, we compute the inverse of ( A ) and multiply by the vector ( B ). The adjugate matrix of ( A ) is:text{adj}(A) = left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right),and the inverse is:A^{-1} = frac{1}{5} left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right).Multiplying ( A^{-1} ) by ( B ):X = A^{-1} B = frac{1}{5} left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right) left(begin{array}{r}5 0 15end{array}right) = left(begin{array}{r}2 1 3end{array}right).Both methods confirm the solution. The final answer is:boxed{left(begin{array}{r} 2 1 3 end{array}right)}.,<think>Okay, so I need to solve this matrix equation where a 3x3 matrix multiplied by some matrix X gives a 3x1 result. The question is asking for X, right? Let me first recall how matrix multiplication works here. Since the result is a 3x1 matrix, X must be a 3x1 matrix as well because the number of columns in the first matrix has to match the number of rows in X. So X is a vector with three entries. Let me write that down as variables. Let me call X = [x; y; z]. Then, when we multiply the given matrix by X, we should get the vector [5; 0; 15]. So, essentially, this is a system of linear equations. Let me write out those equations step by step.First, the given matrix is:3 -1 0-2 1 12 -1 4So multiplying this by X = [x; y; z] gives:First row: 3x - y + 0z = 5Second row: -2x + y + z = 0Third row: 2x - y + 4z = 15So now I have three equations:1. 3x - y = 52. -2x + y + z = 03. 2x - y + 4z = 15Alright, now I need to solve this system of equations. Let me see the best way to approach this. Maybe substitution or elimination. Let's start with the first equation. From equation 1, we can express y in terms of x. Let's do that.From equation 1: 3x - y = 5 ⇒ y = 3x - 5Okay, so now we can substitute y = 3x - 5 into the other equations. Let's plug this into equation 2 and equation 3.Starting with equation 2: -2x + y + z = 0Substitute y: -2x + (3x - 5) + z = 0Simplify: (-2x + 3x) + (-5) + z = 0 ⇒ x - 5 + z = 0 ⇒ x + z = 5So equation 2 becomes: x + z = 5 ⇒ z = 5 - xAlright, so now we have z expressed in terms of x. Let's substitute both y and z into equation 3.Equation 3: 2x - y + 4z = 15Substitute y = 3x - 5 and z = 5 - x:2x - (3x - 5) + 4*(5 - x) = 15Let me compute each term:2x - 3x + 5 + 20 - 4x = 15Combine like terms:(2x - 3x - 4x) + (5 + 20) = 15(-5x) + 25 = 15So, -5x + 25 = 15Subtract 25 from both sides:-5x = -10Divide by -5:x = 2Alright, so we found x = 2. Now, let's substitute back into y and z.First, y = 3x -5. So if x=2:y = 3*2 -5 = 6 -5 = 1Then, z = 5 - x = 5 -2 = 3So, the solution should be x=2, y=1, z=3. Let me check if this satisfies all three equations.First equation: 3x - y = 3*2 -1 = 6 -1 =5. Correct, matches the first entry.Second equation: -2x + y + z = -4 +1 +3 = 0. Correct, that's 0 as given.Third equation: 2x - y +4z = 4 -1 +12 =15. Yes, that's 15. Perfect.So, all equations are satisfied. Therefore, the solution is X = [2; 1; 3].Wait, let me make sure there's no other way this could be approached. Maybe using inverse matrices? Since the coefficient matrix is 3x3, we could compute its inverse and multiply both sides by it. Let me see if that's feasible here.Given that the matrix equation is A*X = B, where A is the 3x3 matrix, X is the vector [x; y; z], and B is [5; 0; 15]. Then, X = A^{-1}*B, provided A is invertible.So, maybe computing the inverse of A is another way. But since we already solved it by substitution and got the answer, maybe computing the inverse is overkill, but just to check, let's verify if the determinant is non-zero.The determinant of A:|A| = 3*(1*4 - (-1)*1) - (-1)*(-2*4 -1*2) + 0*(...)Wait, let's compute it step by step.Using the first row for expansion:det(A) = 3*det(minor of 3) - (-1)*det(minor of -1) + 0*det(minor of 0)Compute minors:Minor of 3 is the submatrix:1 1-1 4Determinant: (1*4) - (1*(-1)) = 4 +1 =5Minor of -1 is:-2 12 4Determinant: (-2*4) - (1*2) = -8 -2 = -10Minor of 0 is irrelevant because multiplied by 0.So, det(A) = 3*5 - (-1)*(-10) +0 = 15 -10 =5So determinant is 5, which is non-zero. Hence, the matrix is invertible. Therefore, the inverse exists. So, if I compute A^{-1} and multiply by B, I should get the same solution.But since we already have the solution, maybe this is not necessary, but just to confirm, let's try to compute the inverse.The inverse of A is (1/det(A)) * adjugate(A). The adjugate is the transpose of the cofactor matrix.First, compute the matrix of minors, then apply the checkerboard of signs, then transpose.Alternatively, for each element, compute the cofactor.But this might take some time. Let's see.First, the cofactor matrix is computed as follows:For each element A[i,j], compute (-1)^{i+j} * det(minor(i,j))So, starting with element (1,1): 3. Its minor is the determinant we already calculated: 5. The cofactor is (+1)^{1+1} *5 =1*5=5Element (1,2): -1. Its minor is the determinant of:-2 12 4Which is (-2*4)-(1*2) = -8 -2= -10. The cofactor is (-1)^{1+2}*(-10)= (-1)^3*(-10)= -1*(-10)=10Element (1,3):0. Minor is:-2 12 -1Determinant: (-2*(-1)) - (1*2)= 2 -2=0. Cofactor is (+1)^{1+3}*0=0Second row:Element (2,1): -2. Minor is:-1 0-1 4Determinant: (-1*4) - (0*(-1))= -4 -0= -4. Cofactor: (-1)^{2+1}*(-4)= (-1)^3*(-4)= -1*(-4)=4Element (2,2):1. Minor:3 02 4Determinant: 3*4 -0*2=12. Cofactor: (-1)^{2+2}*12=1*12=12Element (2,3):1. Minor:3 -12 -1Determinant: 3*(-1) - (-1)*2= -3 +2= -1. Cofactor: (-1)^{2+3}*(-1)= (-1)^5*(-1)= -1*(-1)=1Third row:Element (3,1):2. Minor:-1 01 1Determinant: (-1*1) - (0*1)= -1 -0= -1. Cofactor: (-1)^{3+1}*(-1)=1*(-1)= -1Element (3,2):-1. Minor:3 0-2 1Determinant: 3*1 -0*(-2)=3 -0=3. Cofactor: (-1)^{3+2}*3= (-1)^5*3= -3Element (3,3):4. Minor:3 -1-2 1Determinant: 3*1 - (-1)*(-2)=3 -2=1. Cofactor: (-1)^{3+3}*1=1*1=1Therefore, the cofactor matrix is:[5, 10, 0;4,12,1;-1, -3,1]Then, the adjugate matrix is the transpose of the cofactor matrix:Transposing the cofactor matrix:First column becomes first row:5,4,-1Second column becomes second row:10,12,-3Third column becomes third row:0,1,1So adjugate(A) =5 4 -110 12 -30 1 1Therefore, inverse of A is (1/5)*adjugate(A):So,A^{-1} = (1/5)*[5 4 -1;10 12 -3;0 1 1]Which is:[1, 4/5, -1/5;2, 12/5, -3/5;0, 1/5, 1/5]Now, to compute X = A^{-1} * B, where B is [5;0;15]Multiplying:First row of A^{-1} times B: 1*5 + (4/5)*0 + (-1/5)*15 =5 +0 -3=2Second row:2*5 + (12/5)*0 + (-3/5)*15=10 +0 -9=1Third row:0*5 + (1/5)*0 + (1/5)*15=0 +0 +3=3So X = [2;1;3], which matches the solution we found earlier. So this confirms that the solution is correct.Alternatively, we could use row operations (Gaussian elimination) to solve the system. Let me try that as well to cross-verify.Starting with the augmented matrix:[3 -1 0 |5-2 1 1 |02 -1 4 |15]Let me perform row operations to reduce it.First, let's make the element at (1,1) as 1. But maybe first, let's use the first row to eliminate the elements below it in the first column.First row: 3 -1 0 |5Second row: -2 1 1 |0Third row: 2 -1 4 |15Let's take Row 1: 3 -1 0 |5We can use Row1 to eliminate the first element in Row2 and Row3.For Row2: current first element is -2. Let's compute the multiplier: m = (-2)/3. So add m*Row1 to Row2.Row2_new = Row2 + (2/3)Row1Compute each entry:First entry: -2 + (2/3)*3 = -2 +2=0Second entry:1 + (2/3)*(-1)=1 -2/3=1/3Third entry:1 + (2/3)*0=1Fourth entry:0 + (2/3)*5=10/3So Row2 becomes: 0 1/3 1 |10/3Similarly for Row3: first element is 2. Multiplier m=2/3. Subtract m*Row1 from Row3.Row3_new = Row3 - (2/3)Row1Compute entries:First entry:2 - (2/3)*3=2 -2=0Second entry:-1 - (2/3)*(-1)=-1 +2/3=-1/3Third entry:4 - (2/3)*0=4Fourth entry:15 - (2/3)*5=15 -10/3=35/3So Row3 becomes: 0 -1/3 4 |35/3Now the augmented matrix is:Row1: 3 -1 0 |5Row2: 0 1/3 1 |10/3Row3:0 -1/3 4 |35/3Now, let's focus on the submatrix starting from Row2 and Column2.The element at (2,2) is 1/3. Let's use this to eliminate the element below it in Row3.Row3 has -1/3 in column2. The multiplier here would be (-1/3)/(1/3) = -1. So add Row2 to Row3.Row3_new = Row3 + Row2Compute:0 +0=0-1/3 +1/3=04 +1=535/3 +10/3=45/3=15So Row3 becomes:0 0 5 |15Therefore, the matrix now is:Row1:3 -1 0 |5Row2:0 1/3 1 |10/3Row3:0 0 5 |15Now, we can back-substitute.From Row3: 5z =15 ⇒ z=3From Row2: (1/3)y + z =10/3We know z=3, so:(1/3)y +3=10/3 ⇒ (1/3)y=10/3 -9/3=1/3 ⇒ y=1From Row1:3x - y =5 ⇒ 3x -1=5 ⇒3x=6 ⇒x=2Same solution as before. So all methods converge to x=2, y=1, z=3. Therefore, the solution is correct.Alternatively, maybe using Cramer's rule? Let's try that as well for thoroughness.Cramer's rule says that each variable is the ratio of determinants, where the numerator is the determinant of the matrix formed by replacing the corresponding column with the constants vector.Given the system:A = [3 -1 0;-2 1 1;2 -1 4]B = [5; 0;15]The determinant of A is 5, as computed before.Then,x = det(A_x)/det(A), where A_x is matrix A with first column replaced by B.Similarly for y and z.Compute x:A_x = [5 -1 0;0 1 1;15 -1 4]Compute det(A_x):5*(1*4 - (-1)*1) - (-1)*(0*4 -1*15) +0*(0*(-1) -1*15)=5*(4 +1) - (-1)*(0 -15) +0*(...)=5*5 - (-1)*(-15) +0=25 -15 +0=10Therefore, x=10/5=2Similarly for y:A_y = [3 5 0;-2 0 1;2 15 4]det(A_y):3*(0*4 -1*15) -5*(-2*4 -1*2) +0*(...)=3*(0 -15) -5*(-8 -2) +0=3*(-15) -5*(-10)=-45 +50=5Therefore, y=5/5=1For z:A_z = [3 -1 5;-2 1 0;2 -1 15]det(A_z):3*(1*15 -0*(-1)) -(-1)*(-2*15 -0*2) +5*(-2*(-1) -1*2)=3*(15 -0) -(-1)*(-30 -0) +5*(2 -2)=3*15 - (-1)*(-30) +5*0=45 -30 +0=15Therefore, z=15/5=3So again, x=2, y=1, z=3. Same result.Thus, all methods confirm that the solution is X = [2;1;3].Final AnswerThe solution to the matrix equation is boxed{left(begin{array}{r} 2 1 3 end{array}right)}.</think>To solve the matrix equationleft(begin{array}{rrr}3 & -1 & 0 -2 & 1 & 1 2 & -1 & 4end{array}right) X = left(begin{array}{r}5 0 15end{array}right),we start by expressing the matrix equation as a system of linear equations:1. (3x - y = 5)2. (-2x + y + z = 0)3. (2x - y + 4z = 15)First, solve equation 1 for (y):[ y = 3x - 5 ]Substitute (y = 3x - 5) into equations 2 and 3:2. (-2x + (3x - 5) + z = 0 implies x - 5 + z = 0 implies z = 5 - x)3. (2x - (3x - 5) + 4(5 - x) = 15)Simplify equation 3:[2x - 3x + 5 + 20 - 4x = 15 implies -5x + 25 = 15 implies -5x = -10 implies x = 2]Substitute (x = 2) back into the expressions for (y) and (z):[y = 3(2) - 5 = 1][z = 5 - 2 = 3]Verify the solution (x = 2), (y = 1), (z = 3) in all equations:1. (3(2) - 1 = 5)2. (-2(2) + 1 + 3 = 0)3. (2(2) - 1 + 4(3) = 15)All equations are satisfied. Thus, the solution to the matrix equation isboxed{left(begin{array}{r} 2 1 3 end{array}right)}]

🔑:Solution. 1^{0}. Find the inverse matrix A^{-1}.Calculate the determinant of matrix A:begin{gathered}D=left|begin{array}{rrr}3 & -1 & 0 -2 & 1 & 1 2 & -1 & 4end{array}right|=3 cdot 1 cdot 4+(-1) cdot 1 cdot 2+(-2) cdot(-1) cdot 0- -2 cdot 1 cdot 0-(-1) cdot 1 cdot 3-(-2) cdot(-1) cdot 4=12-2+3-3=5 neq 0end{gathered}Write down all the algebraic complements of the elements of matrix A:begin{gathered}A_{11}=(-1)^{1+1} cdotleft|begin{array}{rr}1 & 1 -1 & 4end{array}right|=5 ; A_{12}=(-1)^{1+2} cdotleft|begin{array}{rr}-2 & 1 2 & 4end{array}right|=10 A_{13}=(-1)^{1+3} cdotleft|begin{array}{rr}-2 & 1 2 & -1end{array}right|=0 ; A_{21}=(-1)^{2+1} cdotleft|begin{array}{cc}-1 & 0 -1 & 4end{array}right|=4 ; A_{22}=(-1)^{2+2} cdotleft|begin{array}{rr}3 & 0 2 & 4end{array}right|=12 A_{23}=(-1)^{2+3} cdotleft|begin{array}{cc}3 & -1 2 & -1end{array}right|=1 ; A_{31}=(-1)^{3+1} cdotleft|begin{array}{rr}-1 & 0 1 & 1end{array}right|=-1 ; A_{32}=(-1)^{3+2} cdotleft|begin{array}{rr}3 & 0 -2 & 1end{array}right|=-3 ; A_{33}^{prime}=(-1)^{3+3} cdotleft|begin{array}{rr}3 & -1 -2 & 1end{array}right|=1end{gathered}Write down the new matrixleft(begin{array}{rrr}5 & 10 & 0 4 & 12 & 1 -1 & -3 & 1end{array}right)and transpose it:left(begin{array}{rrr}5 & 4 & -1 10 & 12 & -3 0 & 1 & 1end{array}right)Considering that 1 / D=1 / 5, write down the inverse matrix:A^{-1}=left(begin{array}{rrr}1 & frac{4}{5} & -frac{1}{5} 2 & frac{12}{5} & -frac{3}{5} 0 & frac{1}{5} & frac{1}{5}end{array}right)2^{0}. We havebegin{aligned}X= & left(begin{array}{rrr}1 & frac{4}{5} & -frac{1}{5} 2 & frac{12}{5} & -frac{3}{5} 0 & frac{1}{5} & frac{1}{5}end{array}right)left(begin{array}{l}5 0 15end{array}right)=left(begin{array}{l}1 cdot 5+frac{4}{5} cdot 0+left(-frac{1}{5}right) cdot 15 2 cdot 5+frac{12}{5} cdot 0+left(-frac{3}{5}right) cdot 15 0 cdot 5+frac{1}{5} cdot 0+frac{1}{5} cdot 15end{array}right)=left(begin{array}{l}2 1 3end{array}right) & 3^{0} . text { Thus, }left(begin{array}{l}x_{1} x_{2} x_{3}end{array}right)=left(begin{array}{l}2 1 3end{array}right), text { i.e., } x_{1}=2, x_{2}=1, x_{3}=3end{aligned}62-65. Solve the matrix equations:

🧠:[<think>Okay, let's see. I need to find the angle between two refracted rays that are mutually perpendicular and incident on a horizontal glass surface. The refractive index of the glass is 1.5, and the angle of refraction for the first ray is 25 degrees. Hmm.First, let me visualize this. There are two rays coming from air (which has a refractive index of 1) into glass. They are mutually perpendicular, so their original directions in air form a 90-degree angle with each other. Both are in the same vertical plane, which I guess means they're both in the plane of incidence, but since they're mutually perpendicular, maybe one is in the plane and the other is perpendicular? Wait, no, the problem says they're in the same vertical plane. So both rays are in the same vertical plane, and they are mutually perpendicular before refraction. So their incident directions are perpendicular to each other.Now, after refraction, they enter the glass, and I need to find the angle between the two refracted rays. Given that one of the refracted angles is 25 degrees. Let's denote the first ray as Ray 1 with refraction angle β1 = 25 degrees. Then the second ray, Ray 2, which was perpendicular to Ray 1 in air, will have some other angle of refraction β2. The angle between the two refracted rays will be the difference between β1 and β2 or maybe 180 minus that? Wait, no, since they are in the same plane, it's just the absolute difference if they are on the same side of the normal. Wait, actually, need to think more carefully.Wait, the two incident rays are mutually perpendicular. So, if one is incident at angle θ1, the other is incident at angle θ2 such that θ1 + θ2 = 90 degrees? Because they're perpendicular. Wait, maybe not exactly, because in 3D space, two lines can be perpendicular but have different relationships with the surface. Wait, but since they are in the same vertical plane, their directions are perpendicular in that plane. Hmm.Wait, actually, if two vectors are perpendicular in a plane, their angles with respect to a reference line (like the normal) would satisfy some trigonometric condition. Let me think.Suppose the first incident ray makes an angle θ1 with the normal. The second incident ray is perpendicular to the first one in the plane. So the direction of the second ray would make an angle θ2 with the normal such that the angle between the two incident rays is 90 degrees. The angle between two lines in a plane can be found using the difference in their angles with respect to a reference. If the first ray is at θ1 from the normal, and the second is at θ2 from the normal, then the angle between the two incident rays is |θ1 - θ2|. But the problem states that they are mutually perpendicular, so |θ1 - θ2| = 90 degrees. Wait, but angles of incidence are measured from the normal, so if θ1 is the angle of incidence for Ray 1, then the angle between Ray 1 and the surface is 90 - θ1. Similarly for Ray 2, it's 90 - θ2. Then the angle between the two incident rays would be (90 - θ1) + θ2, if they are on opposite sides of the normal. Wait, this is getting confusing.Alternatively, maybe since they are in the same plane and mutually perpendicular, the angle between their directions is 90 degrees. So the angle between their direction vectors is 90 degrees. Let's model this.Assuming the normal is vertical (since the glass surface is horizontal), then the incident rays are in the vertical plane. Let me define coordinate system: let the normal be along the z-axis, and the vertical plane is the x-z plane. The two incident rays are in this plane and are perpendicular to each other.The direction vectors of the incident rays can be represented as vectors in 3D. Since they are in the x-z plane, their y-components are zero. Let the first incident ray have direction vector (a, 0, b), and the second incident ray have direction vector (c, 0, d). Since they are perpendicular, their dot product is zero: a*c + 0*0 + b*d = 0 => a*c + b*d = 0.But they are both coming from air into glass, so their direction vectors are pointing towards the surface. Assuming the glass is below the air, so the z-axis points upwards, then the incident rays are going downward, so their z-components are negative. Wait, maybe this is complicating.Alternatively, think in terms of angles with respect to the normal. For Ray 1, angle of incidence is θ1, so its direction vector makes an angle θ1 with the normal (z-axis). So in x-z plane, the direction vector can be represented as (sinθ1, 0, -cosθ1) because it's going downward. Similarly, Ray 2 is perpendicular to Ray 1, so the dot product between their direction vectors should be zero:(sinθ1)(sinθ2) + (-cosθ1)(-cosθ2) = 0Wait, if Ray 2 has angle of incidence θ2, its direction vector is (sinθ2, 0, -cosθ2). Then the dot product is sinθ1 sinθ2 + cosθ1 cosθ2 = 0But sinθ1 sinθ2 + cosθ1 cosθ2 = cos(θ1 - θ2) = 0Therefore, cos(θ1 - θ2) = 0 => θ1 - θ2 = ±90 degreesBut angles of incidence θ1 and θ2 are between 0 and 90 degrees, so θ1 - θ2 = 90 degrees is impossible because θ1 would have to be more than 90. Similarly θ2 - θ1 = 90 is impossible. Therefore, this approach might not be correct.Wait, maybe the problem states that the two incident rays are mutually perpendicular, meaning their direction vectors are perpendicular. But given that they are in the same vertical plane, perhaps the angle between their projections onto the horizontal plane is 90 degrees? Wait, no, they are in a vertical plane, so their projections onto the horizontal plane would both be along the same line, so the angle between them would be zero. Hmm, maybe I need to think differently.Wait, maybe "mutually perpendicular" means that the planes of incidence are perpendicular? But the problem says they are in the same vertical plane. Hmm, the problem states: "Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface."So both rays are in the same vertical plane, and they are mutually perpendicular. So their direction vectors are perpendicular within that vertical plane. So, in the vertical plane, if one ray is going in some direction, the other is going in a direction perpendicular to that within the plane.So, for instance, in the vertical x-z plane, one ray could be going along the x-axis, and the other along the z-axis, but since they are propagating towards the glass surface, which is horizontal, say the glass is at z=0, and the rays are coming from z>0.Wait, but if the surface is horizontal, the normal is vertical (along z-axis). So if a ray is propagating in the vertical plane, its direction can be decomposed into x and z components.If two rays are mutually perpendicular, their direction vectors must have a dot product of zero. Let’s assume Ray 1 has direction vector (a, 0, b), and Ray 2 has direction vector (c, 0, d). Since they are in the same vertical plane (x-z), the y-component is zero. The dot product is a*c + b*d = 0.But both rays are propagating towards the surface, so their z-components (assuming z is the vertical direction) are negative (if the surface is at z=0 and they come from above). So b and d are negative. Let's write this as (sinθ1, 0, -cosθ1) and (sinθ2, 0, -cosθ2), where θ1 and θ2 are the angles of incidence for each ray.Wait, no. The direction vector of a ray with angle θ from the normal would have a z-component of cosθ and the x-component of sinθ. But since it's propagating towards the surface, the z-component should be negative. So direction vector is (sinθ, 0, -cosθ). Similarly for the other ray.So the dot product between the two direction vectors is:(sinθ1)(sinθ2) + (0)(0) + (-cosθ1)(-cosθ2) = sinθ1 sinθ2 + cosθ1 cosθ2 = cos(θ1 - θ2)Since the rays are mutually perpendicular, their direction vectors are perpendicular, so the dot product is zero:cos(θ1 - θ2) = 0Therefore, θ1 - θ2 = ±90 degrees + 360k, but since θ1 and θ2 are angles of incidence between 0 and 90 degrees, the only possibility is θ1 - θ2 = ±90 degrees. But since both angles are between 0 and 90, θ1 - θ2 = 90 is impossible because θ1 would have to be greater than 90. Similarly θ2 - θ1 = 90 is impossible. So this seems contradictory.Wait, this suggests that two rays in the same vertical plane cannot be mutually perpendicular if they are incident on a horizontal surface. But the problem states that they are. Therefore, maybe my coordinate system is wrong.Alternatively, perhaps "mutually perpendicular" refers to their polarization? No, the problem mentions rays, which are geometric optics, not physical optics with polarization.Wait, perhaps the problem is that the two incident rays are perpendicular to each other outside the glass, meaning before refraction. So their incident directions are perpendicular. Then after refraction, they bend, and we need to find the angle between the refracted rays.But according to Snell's law, each ray refracts into the glass. The first ray has angle of refraction β1 = 25 degrees, so using Snell's law, we can find its angle of incidence θ1. Then, the second ray is perpendicular to the first in air, so θ2 is related to θ1 such that their incident directions are perpendicular. Then using Snell's law, we can find β2, and then compute the angle between β1 and β2.Wait, this seems plausible.Let me try this approach.Given:n_air = 1.0n_glass = 1.5For Ray 1: angle of refraction β1 = 25 degrees.Using Snell's law: n_air sinθ1 = n_glass sinβ1Thus, sinθ1 = n_glass / n_air * sinβ1 = 1.5 * sin(25°)Compute sin(25°): approximately 0.4226Thus, sinθ1 ≈ 1.5 * 0.4226 ≈ 0.6339Therefore, θ1 ≈ arcsin(0.6339) ≈ 39.3 degreesSo angle of incidence for Ray 1 is approximately 39.3 degrees.Now, the two incident rays are mutually perpendicular. So the angle between their directions is 90 degrees. Since they are in the same vertical plane, the angle between their directions can be related to their angles of incidence.Wait, how?If Ray 1 is incident at θ1 = 39.3 degrees from the normal, then its direction makes an angle of θ1 with the normal. The direction vector in air is (sinθ1, 0, -cosθ1) as before.Similarly, Ray 2 is perpendicular to Ray 1, so their direction vectors must have a dot product of zero.Let Ray 2 have direction vector (sinθ2, 0, -cosθ2). Then:(sinθ1)(sinθ2) + (-cosθ1)(-cosθ2) = 0Which simplifies to:sinθ1 sinθ2 + cosθ1 cosθ2 = cos(θ1 - θ2) = 0Therefore, θ1 - θ2 = ±90 degrees. But θ1 is 39.3 degrees, so θ2 = θ1 ±90 degrees. But θ2 must be between 0 and 90 degrees, so θ2 = 39.3 - 90 = -50.7 degrees, which is not possible, or θ2 = 39.3 + 90 = 129.3 degrees, also impossible. Therefore, this approach leads to a contradiction.Wait, so maybe the problem is not that their angles of incidence differ by 90 degrees, but that their actual directions in space are perpendicular?Alternatively, perhaps the two incident rays are perpendicular outside the glass, meaning that the angle between their paths in air is 90 degrees, which is different from their angles with respect to the normal.So if Ray 1 is incident at angle θ1, its direction makes angle θ1 with the normal. The angle between the direction of Ray 1 and the surface is 90 - θ1. Similarly for Ray 2, angle with the surface is 90 - θ2. If the two rays are mutually perpendicular, then the angle between them is 90 degrees. So (90 - θ1) + (90 - θ2) = 90? Wait, that would be 180 - θ1 - θ2 = 90 => θ1 + θ2 = 90.But is this correct?If two lines are in the same plane and intersect at a point, the angle between them is the difference in their angles with respect to a reference line. If one is at angle α and the other at angle β from the reference, then the angle between them is |α - β|. But if they are on opposite sides of the reference line, it could be α + β.In our case, the reference line is the surface. So if Ray 1 makes an angle α = 90 - θ1 with the surface (since θ1 is the angle with the normal), and Ray 2 makes an angle β = 90 - θ2 with the surface. If both rays are on the same side of the normal, then the angle between them is |α - β| = |(90 - θ1) - (90 - θ2)| = |θ2 - θ1|. If they are on opposite sides, then the angle between them would be α + β = (90 - θ1) + (90 - θ2) = 180 - θ1 - θ2.But the problem says they are mutually perpendicular, so angle between them is 90 degrees. Therefore, either |θ2 - θ1| = 90 or 180 - θ1 - θ2 = 90. Since θ1 and θ2 are both between 0 and 90, 180 - θ1 - θ2 = 90 implies θ1 + θ2 = 90. That seems possible. The other case |θ2 - θ1| = 90 is impossible because θ1 and θ2 are both less than 90.Therefore, θ1 + θ2 = 90 degrees.So from earlier, θ1 ≈ 39.3 degrees, so θ2 = 90 - 39.3 ≈ 50.7 degrees.Therefore, angle of incidence for Ray 2 is approximately 50.7 degrees.Then, using Snell's law for Ray 2:n_air sinθ2 = n_glass sinβ2Therefore, sinβ2 = (n_air / n_glass) sinθ2 = (1 / 1.5) sin(50.7°)Calculate sin(50.7°): approximately 0.774Thus, sinβ2 ≈ (1 / 1.5) * 0.774 ≈ 0.516Therefore, β2 ≈ arcsin(0.516) ≈ 31 degreesSo the angle of refraction for Ray 2 is approximately 31 degrees.Now, we need to find the angle between the two refracted rays in the glass. The refracted rays have angles β1 = 25 degrees and β2 ≈ 31 degrees with respect to the normal.Since both refracted rays are on the same side of the normal (below the surface), the angle between them is the difference between their angles with respect to the normal.Therefore, angle between refracted rays = β2 - β1 ≈ 31 - 25 = 6 degrees.Wait, but that seems too small. Let me check the calculations again.Wait, first, θ1 was found to be arcsin(1.5 * sin25°). Wait, no: Snell's law is n1 sinθ1 = n2 sinβ1. So here, n1 = 1 (air), n2 = 1.5 (glass). So sinθ1 = (n2 / n1) sinβ1 = 1.5 * sin25°, which is approximately 1.5 * 0.4226 ≈ 0.6339. So θ1 ≈ 39.3 degrees. Then, since θ1 + θ2 = 90 degrees (from the angle between the incident rays), θ2 = 90 - 39.3 ≈ 50.7 degrees.Then, for Ray 2, sinβ2 = (1 / 1.5) * sinθ2 = (1 / 1.5) * sin50.7°. sin50.7 is approx 0.774, so sinβ2 ≈ 0.774 / 1.5 ≈ 0.516, so β2 ≈ 31 degrees.Therefore, the two refracted rays make angles of 25 degrees and 31 degrees with the normal, both on the same side. Thus, the angle between them is 31 - 25 = 6 degrees.But wait, 6 degrees seems quite small. Is this correct? Let me think again.Alternatively, maybe the angle between the refracted rays is not just the difference in their angles with the normal. Because both are in the same plane, but depending on their directions, the angle between them would be the difference if they are on the same side of the normal. For example, if one is refracted at 25 degrees to the left of the normal and the other at 31 degrees to the right, the angle between them would be 25 + 31 = 56 degrees. But in this case, the problem says they are in the same vertical plane, so both refracted rays are in the same vertical plane. If the two incident rays are in the same vertical plane and mutually perpendicular, then their refracted rays should also be in that same vertical plane. Therefore, if one is refracted at 25 degrees on one side of the normal and the other at 31 degrees on the same side, the angle between them is 6 degrees. But if they are on opposite sides, it would be 25 + 31 = 56 degrees. Wait, but the problem states that they are propagating in the same vertical plane. If the incident rays are in the same vertical plane and mutually perpendicular, their refracted rays should also be in the same vertical plane, but on the same side of the normal (since they are both going into the glass). Therefore, their refracted angles are measured from the normal in the same direction, so the angle between the refracted rays would be the difference between β2 and β1, which is 6 degrees. However, this seems counterintuitive because the original incident rays were perpendicular, but after refraction, they are only 6 degrees apart. That seems like a very small angle. Maybe my approach is wrong.Wait, maybe the confusion is arising from how the angle between the two rays is calculated. Let me model their direction vectors.In the glass, the refracted rays have angles β1 and β2 with respect to the normal (z-axis). Assuming the vertical plane is x-z, with the normal along z-axis. The direction vectors of the refracted rays would be:For Ray 1: (sinβ1, 0, -cosβ1) because it's going downward (negative z) and making angle β1 with the normal.For Ray 2: (sinβ2, 0, -cosβ2)Then, the angle between these two vectors can be found using the dot product formula:cosφ = (sinβ1 sinβ2 + cosβ1 cosβ2) / (|v1||v2|)Since the vectors are unit vectors (since direction vectors are normalized), the denominator is 1.Therefore, cosφ = sinβ1 sinβ2 + cosβ1 cosβ2 = cos(β1 - β2)Therefore, φ = |β1 - β2|Wait, that's interesting. So the angle between the two refracted rays is equal to the difference between their angles with the normal. So if β1 = 25°, β2 = 31°, then the angle between them is 6°. So that is correct.But 6 degrees seems small, but according to the calculation, it's correct. Alternatively, maybe I messed up the calculation for β2.Wait, let's recalculate with more precise numbers.First, compute θ1:sinθ1 = 1.5 * sin25°sin25° ≈ 0.42261826174So sinθ1 ≈ 1.5 * 0.42261826174 ≈ 0.63392739261θ1 ≈ arcsin(0.63392739261) ≈ 39.328 degreesTherefore, θ1 ≈ 39.328°Then θ2 = 90° - θ1 ≈ 90 - 39.328 ≈ 50.672°Compute sinθ2 = sin(50.672°) ≈ let's calculate 50 degrees is sin(50) ≈ 0.7660, 50.672°:Difference is 0.672°, so sin(50.672) ≈ sin(50 + 0.672) ≈ sin50 + 0.672*(π/180)*cos50 ≈ 0.7660 + 0.0117*0.6428 ≈ 0.7660 + 0.0075 ≈ 0.7735Therefore, sinθ2 ≈ 0.7735Then sinβ2 = (1/1.5)*0.7735 ≈ 0.5157Therefore, β2 ≈ arcsin(0.5157) ≈ 31.0 degreesTherefore, β2 ≈ 31.0 degreesTherefore, the angle between the refracted rays is 31.0 - 25 = 6.0 degrees.But wait, is this correct? The two refracted rays are in the same vertical plane, one at 25 degrees from the normal, the other at 31 degrees from the normal. Since both are on the same side of the normal, the angle between them is 6 degrees. Hmmm.Alternatively, maybe the angle is measured as the angle between their paths in the glass, which would indeed be 6 degrees. But this seems very small, but perhaps correct. Let's check with an example.Suppose two rays incident along the normal (θ=0) and another at θ=90 (grazing incidence). The first would refract at 0 degrees (no bending), the second would have a critical angle. Wait, but θ=90 is not possible for refraction as it would require sinβ = (1/1.5)*sin90 ≈ 0.666, so β≈41.8 degrees. The angle between them would be 41.8 - 0 = 41.8 degrees.But in our case, the two incident rays are not that extreme. One is at 39.3 degrees, the other at 50.7 degrees incidence. Their refracted angles are 25 and 31 degrees. The difference is 6 degrees.Alternatively, maybe my initial assumption that θ1 + θ2 = 90 degrees is incorrect.Wait, let's revisit the angle between the incident rays. The problem states that the two incident rays are mutually perpendicular. If they are in the same vertical plane, and their direction vectors are perpendicular, then the dot product of their direction vectors must be zero. The direction vectors in air are:Ray 1: (sinθ1, 0, -cosθ1)Ray 2: (sinθ2, 0, -cosθ2)Dot product: sinθ1 sinθ2 + cosθ1 cosθ2 = cos(θ1 - θ2) = 0Therefore, θ1 - θ2 = ±90 degrees, but since θ1 and θ2 are both positive and less than 90, this is impossible. Therefore, this suggests that two rays in the same vertical plane cannot have their direction vectors perpendicular unless one of them is going upward (positive z) and the other downward. But in our case, both are incident rays coming from air into glass, so they are both going downward (negative z direction). Therefore, their z-components are both negative. Hence, the dot product sinθ1 sinθ2 + cosθ1 cosθ2 = cos(θ1 - θ2) = 0.But this is impossible unless θ1 - θ2 = ±90 degrees, which cannot happen. Therefore, there must be something wrong with the problem statement or my interpretation.Wait, but the problem says "Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface." So the problem says they are mutually perpendicular. Therefore, perhaps the error is in the assumption that the angle between the direction vectors is 90 degrees. Maybe instead, the angle between their wave vectors is 90 degrees? But wave vectors in isotropic media are in the same direction as the rays. So same thing.Alternatively, perhaps the rays are perpendicular after accounting for the reflection? Wait, no, the problem states they are mutually perpendicular before incidence.Alternatively, maybe "mutually perpendicular" refers to their planes of polarization? But the problem doesn't mention polarization.Alternatively, maybe they are perpendicular in 3D space, but both in the same vertical plane. Wait, but in a single vertical plane, two lines can only be perpendicular if one is vertical and the other horizontal. For example, one along the vertical (normal) direction and the other along the horizontal direction. But in that case, the vertical ray would have θ1 = 0 degrees (along the normal), and the horizontal ray would have θ2 = 90 degrees (grazing incidence). But θ2 = 90 degrees is not possible for refraction, as sinβ2 would be (1/1.5)*sin90 = 2/3, so β2 = arcsin(2/3) ≈ 41.81 degrees. Then, the angle between the refracted rays would be 41.81 - 0 = 41.81 degrees. But in this case, the problem states that one of the angles of refraction is 25 degrees, not 41.81. So this example does not match the problem's given data.Wait, but in our problem, one of the refracted angles is given as 25 degrees, so we cannot have θ2 = 90 degrees. Therefore, my previous assumption that θ1 + θ2 = 90 degrees must be incorrect.This suggests that there is a contradiction in the problem statement or my interpretation. Because according to the dot product, two rays in the same plane with direction vectors perpendicular must have θ1 - θ2 = ±90 degrees, which is impossible. Therefore, maybe the problem is intended to have the two incident rays being perpendicular outside the plane? But the problem says they are in the same vertical plane.Wait, maybe the rays are not both in the plane of incidence. Wait, the problem says "Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface."So both rays are in the same vertical plane, which is the plane of incidence for both. Then, being mutually perpendicular in that plane. So their direction vectors are perpendicular. As we saw earlier, this leads to a requirement that cos(θ1 - θ2) = 0, which is impossible. Therefore, the problem might have a typo, or perhaps "mutually perpendicular" refers to something else.Alternatively, maybe the two rays are polarized perpendicularly, but the problem doesn't mention polarization.Alternatively, maybe the problem means that the two rays are incident such that their planes of incidence are perpendicular. But the problem states they are in the same vertical plane.This is confusing. Given that the problem is from an exam, perhaps the intended solution is as I did earlier, assuming θ1 + θ2 = 90 degrees, leading to angle between refracted rays being 6 degrees. But I need to check again.Wait, if θ1 + θ2 = 90 degrees, then θ2 = 90 - θ1. Then, using Snell's law for both rays:For Ray 1: sinθ1 = 1.5 sinβ1 (given β1 = 25°, so θ1 ≈ 39.3°)For Ray 2: sinθ2 = 1.5 sinβ2, but θ2 = 90 - θ1, so sinθ2 = sin(90 - θ1) = cosθ1Therefore, cosθ1 = 1.5 sinβ2 => sinβ2 = (cosθ1)/1.5Since θ1 ≈ 39.3°, cosθ1 ≈ cos(39.3°) ≈ 0.774Therefore, sinβ2 ≈ 0.774 / 1.5 ≈ 0.516 => β2 ≈ 31°, as before.Then angle between refracted rays is 31° - 25° = 6°. Therefore, this is consistent.But the problem is, why is θ1 + θ2 = 90°? Because if the two incident rays are mutually perpendicular, the angle between them is 90°, which in this case, since they are in the same plane, and angles with the surface are 90 - θ1 and 90 - θ2, the angle between the incident rays is (90 - θ1) + (90 - θ2) = 180 - θ1 - θ2. If this is 90°, then θ1 + θ2 = 90°, as previously thought.So if the angle between the incident rays is 90°, then θ1 + θ2 = 90°, leading to the angle between refracted rays being 6°, as calculated.Therefore, even though the mathematical condition from the dot product suggests it's impossible, perhaps the problem is using a different definition where the angle between the incident rays is 90°, leading to θ1 + θ2 = 90°, hence angle between refracted rays 6°. Given that this is a problem from an exam, this is likely the intended solution.Therefore, the answer is 6 degrees. But let me check units. The problem says "angle between the refracted rays," so in degrees, 6°. But let me verify once more.Alternatively, perhaps the angle between the refracted rays is 180° - 6° = 174°, but that doesn't make sense because they are both on the same side of the normal.No, if both refracted rays are below the surface (in the glass) and in the same vertical plane, then the angle between them is simply the difference between their angles with respect to the normal. So 31° - 25° = 6°.Yes, that seems to be the case. So the answer is 6 degrees.But to confirm, let's take an example with known angles.Suppose θ1 = 0°, then β1 = 0°. The other incident ray would have θ2 = 90° (since θ1 + θ2 = 90°), but θ2 = 90° is grazing incidence. Then sinβ2 = (1/1.5) sin90° = 2/3, β2 ≈ 41.81°. Then angle between refracted rays is 41.81° - 0° = 41.81°, which makes sense.Another example: θ1 = 45°, so θ2 = 45°, but wait, θ1 + θ2 = 90°, so if θ1 = 30°, θ2 = 60°. Then sinβ1 = (1/1.5) sin30° = (1/1.5)*0.5 ≈ 0.333, so β1 ≈ 19.47°. Similarly, sinβ2 = (1/1.5) sin60° ≈ (1/1.5)*(√3/2) ≈ 0.577, so β2 ≈ 35.26°. The angle between them is 35.26 - 19.47 ≈ 15.79°, which is roughly 15.8°, which is the difference. So this method holds.Therefore, in our original problem, angle between refracted rays is 6°, so the answer is 6 degrees. Therefore, boxed answer is boxed{6^circ}Final AnswerThe angle between the refracted rays is boxed{6^circ}.</think>Given two mutually perpendicular rays propagating in the same vertical plane and falling onto a horizontal glass surface with a refractive index ( n = 1.5 ), we need to determine the angle between the refracted rays. The angle of refraction for the first ray is given as ( beta_1 = 25^circ ).1. Calculate the angle of incidence for the first ray using Snell's Law: [ sin theta_1 = n sin beta_1 ] Substituting the values: [ sin theta_1 = 1.5 sin 25^circ approx 1.5 times 0.4226 approx 0.6339 ] [ theta_1 approx arcsin(0.6339) approx 39.3^circ ]2. Determine the angle of incidence for the second ray: Since the incident rays are mutually perpendicular, the sum of their angles with respect to the normal is ( 90^circ ): [ theta_1 + theta_2 = 90^circ ] Substituting ( theta_1 approx 39.3^circ ): [ theta_2 = 90^circ - 39.3^circ approx 50.7^circ ]3. Calculate the angle of refraction for the second ray using Snell's Law: [ sin beta_2 = frac{sin theta_2}{n} ] Substituting the values: [ sin beta_2 = frac{sin 50.7^circ}{1.5} approx frac{0.774}{1.5} approx 0.516 ] [ beta_2 approx arcsin(0.516) approx 31^circ ]4. Determine the angle between the refracted rays: The angle between the refracted rays is the absolute difference between their angles of refraction: [ text{Angle between refracted rays} = |beta_2 - beta_1| = |31^circ - 25^circ| = 6^circ ]Thus, the angle between the refracted rays is (boxed{6^circ}).,<think>Okay, let's try to tackle this problem. So, we have two mutually perpendicular rays hitting a horizontal glass surface. The glass has a refractive index of 1.5, and the angle of refraction for the first ray is 25 degrees. We need to find the angle between the two refracted rays.First, let me visualize the scenario. Two rays are coming from air (which has a refractive index of 1) and hitting a glass surface. These two rays are perpendicular to each other, meaning the angle between their incident directions is 90 degrees. They're in the same vertical plane, so they're not diverging in different planes. The glass surface is horizontal, so the normal to the surface is vertical.The first ray has a refracted angle of 25 degrees. Since refraction occurs at the surface, we can use Snell's Law here. Snell's Law states that n1 * sin(theta1) = n2 * sin(theta2), where theta1 is the angle of incidence and theta2 is the angle of refraction. For the first ray, n1 is 1 (air), n2 is 1.5 (glass), and theta2 is 25 degrees. So, we can find the angle of incidence for the first ray.Let me calculate that. So, rearranging Snell's Law: sin(theta1) = (n2 / n1) * sin(theta2). Plugging in the numbers: sin(theta1) = 1.5 * sin(25°). Wait, hold on, that can't be right. Wait, no, Snell's Law is n1 * sin(theta1) = n2 * sin(theta2). Therefore, sin(theta1) = (n2 / n1) * sin(theta2). But n2 is 1.5 and n1 is 1, so sin(theta1) = 1.5 * sin(25°). Let me compute sin(25°) first. Sin(25°) is approximately 0.4226. Multiplying that by 1.5 gives 0.6339. But the sine of an angle can't be more than 1. Wait a minute, that's a problem. 0.6339 is less than 1, so it's okay. Wait, no, sin(theta1) = 1.5 * sin(25°) ≈ 1.5 * 0.4226 ≈ 0.6339. So theta1 is arcsin(0.6339) ≈ 39.3 degrees. So the angle of incidence for the first ray is approximately 39.3 degrees.Okay, so the first ray is incident at 39.3 degrees from the normal and refracts to 25 degrees in the glass. Now, the second ray is perpendicular to the first ray. Since the two rays are mutually perpendicular, the angle between their directions of propagation in air is 90 degrees. So, if the first ray is incident at 39.3 degrees from the normal, the second ray, being perpendicular, must have an angle of incidence such that the angle between the two incident rays is 90 degrees.Wait, how exactly are the rays arranged? Since they are mutually perpendicular and in the same vertical plane, their directions must form a right angle. So, if the first ray is coming in at some angle to the normal, the second ray is coming in such a way that the angle between the two incident rays is 90 degrees.Let me think. Let's denote the first ray as having an angle of incidence theta1, and the second ray as theta1'. The angle between the two incident rays is 90 degrees. Since both rays are in the same vertical plane, which contains the normal, the two incident rays are in this plane. The angle between them is 90 degrees. So, if the first ray makes an angle theta1 with the normal, the second ray must make an angle of 90 degrees - theta1 with the normal? Wait, no. Because if the first ray is at theta1 from the normal, then its direction is theta1 from the normal. The second ray is perpendicular to the first ray. So, the angle between their directions is 90 degrees. But in a plane, if one direction is theta1 from the normal, the other direction should be such that the angle between them is 90 degrees. So, in other words, if the first ray is coming in at angle theta1 to the normal, the second ray is coming in at some angle theta1' to the normal, such that the angle between the two incident rays is 90 degrees. So, in terms of vectors, the two incident directions would form a 90-degree angle. But how do we relate theta1 and theta1'? Let's consider that the normal is vertical, so the surface is horizontal. The first ray is in the vertical plane, making an angle theta1 with the vertical (normal). The second ray is also in the same vertical plane but direction is such that the angle between the two rays is 90 degrees. Wait, this might be a bit tricky. Maybe using coordinates can help.Let's set up a coordinate system where the normal is along the z-axis, and the vertical plane is the x-z plane. The first ray is in the x-z plane, making an angle theta1 with the z-axis (normal). Its direction vector can be represented as (sin(theta1), 0, cos(theta1)). The second ray is perpendicular to the first ray, so their direction vectors must have a dot product of zero. Since they are in the same x-z plane, the second ray's direction vector would be (sin(theta1'), 0, cos(theta1')). The dot product between the two direction vectors is sin(theta1) * sin(theta1') + cos(theta1) * cos(theta1') = cos(theta1 - theta1') by the cosine difference identity. Wait, but since the angle between the two vectors is 90 degrees, their dot product should be zero. Therefore, cos(theta1 - theta1') = 0. That implies theta1 - theta1' = ±90°, but since theta1 and theta1' are angles from the normal (z-axis) in the same plane, theta1' = theta1 ± 90°. But angles of incidence can't be more than 90°, so theta1' = 90° - theta1? Wait, no. Wait, if theta1 - theta1' = ±90°, but theta1' must be a positive angle less than 90°, since it's an angle of incidence. So, maybe theta1' = theta1 + 90°, but that would be more than 90°, which is not possible. Alternatively, theta1' = theta1 - 90°, but that would be negative. Hmm, this seems conflicting.Alternatively, perhaps the angle between the two incident rays is 90°, so the angle between their direction vectors is 90°, which is different from their angles with the normal. So, if the first ray has an angle theta1 with the normal, the second ray is in the same plane, and the angle between the two rays is 90°, so using vector approach.Let me denote the direction vectors. Let’s take the normal as the z-axis. Then, the first ray's direction vector in air is (sin(theta1), 0, cos(theta1)). The second ray's direction vector is (sin(theta1'), 0, cos(theta1')). The dot product between these two vectors should be zero because they are perpendicular.So:sin(theta1) * sin(theta1') + cos(theta1) * cos(theta1') = 0Using the cosine of difference identity, this is cos(theta1 - theta1') = 0Therefore, theta1 - theta1' = ±90° + 360°k, where k is an integer. But since theta1 and theta1' are both angles of incidence (between 0° and 90°), the only possible solution is theta1 - theta1' = ±90°, but this would require one of the angles to be greater than 90°, which is not possible. Hmm, this seems like a problem. Wait, maybe my coordinate system is wrong.Alternatively, perhaps the two rays are not both in the same vertical plane? Wait, the problem states they are propagating in the same vertical plane. So, they are both in the x-z plane, for example. Then their direction vectors are in the x-z plane. The angle between them is 90 degrees. So, their direction vectors are perpendicular. But how can two vectors in the same plane be perpendicular? For example, in the x-z plane, a vector along the x-axis is (1, 0, 0), and a vector along the z-axis is (0, 0, 1). These are perpendicular. Similarly, a vector at some angle theta1 and another at theta1' such that their dot product is zero.Wait, but in our case, both rays are going towards the surface, so their z-components are positive (assuming the light is coming from above). Wait, no, if they are incident on the surface, the direction vectors would be pointing towards the surface, so the z-component is negative if the surface is at z=0. Wait, maybe I should clarify.Actually, the direction of propagation is towards the surface, so the direction vectors would be pointing towards the negative z-axis. So, for the first ray, direction vector is (sin(theta1), 0, -cos(theta1)), because it's moving downward towards the surface. Similarly, the second ray would have a direction vector (sin(theta1'), 0, -cos(theta1')). Then the dot product between these two vectors is sin(theta1)sin(theta1') + cos(theta1)cos(theta1') = cos(theta1 - theta1'). Wait, same result. So setting this equal to zero, since they are perpendicular, cos(theta1 - theta1') = 0, which again gives theta1 - theta1' = ±90°, which is impossible if both angles are between 0° and 90°. So, there must be a mistake here.Wait, but perhaps the problem is that the two rays are in the same vertical plane but are perpendicular in their directions of propagation, not their direction vectors? Wait, no. The problem states "mutually perpendicular rays, propagating in the same vertical plane". So their directions of propagation are perpendicular. So, their direction vectors are perpendicular. But in a single plane, two vectors can be perpendicular only if one is along the plane and the other is... Wait, no. In a plane, two vectors can be perpendicular. For example, in the x-z plane, a vector along the x-axis and a vector along the z-axis are perpendicular. Similarly, a vector at 45 degrees in x-z plane can be perpendicular to another vector at -45 degrees, but that's in 3D space. Wait, but in 2D plane, two vectors can be perpendicular. For example, (1, 0) and (0, 1) in the x-y plane. But in our case, the plane is x-z, so (1, 0, 0) and (0, 0, 1) are perpendicular.Wait, but if the rays are in the same vertical plane, they can have components in both x and z directions. So, if one ray is going along (sin(theta1), 0, -cos(theta1)), then the other ray, which is perpendicular, would have a direction vector (sin(theta1'), 0, -cos(theta1')) such that their dot product is zero. So:sin(theta1)sin(theta1') + cos(theta1)cos(theta1') = 0Which is equivalent to cos(theta1 - theta1') = 0, as before. Therefore, theta1 - theta1' = ±90°, but since both theta1 and theta1' are between 0° and 90°, theta1' would have to be theta1 - 90° or theta1 + 90°, but both would result in theta1' outside the 0-90° range. Therefore, this seems impossible. Therefore, maybe my approach is wrong.Wait, but the problem says the two rays are mutually perpendicular. Maybe it's not their direction vectors that are perpendicular, but their electric field vectors or something else? No, the problem states "mutually perpendicular rays", so rays here refer to the direction of propagation. So their direction vectors should be perpendicular. But in the same vertical plane. Hmm.Wait, but in three-dimensional space, two rays can be perpendicular even if they are in the same plane. Wait, no. In three-dimensional space, two vectors in the same plane can be perpendicular, but if the plane is 2D, like the x-z plane, then yes. For example, in the x-z plane, the vectors (1, 0, 1) and (1, 0, -1) are perpendicular because their dot product is 1*1 + 0*0 + 1*(-1) = 0. Wait, but those vectors are (1,0,1) and (1,0,-1), which are in the x-z plane. So yes, they can be perpendicular. So perhaps in our case, the two rays have direction vectors that are in the x-z plane and perpendicular.So, let's consider the first ray with direction vector (a, 0, b), and the second ray with direction vector (c, 0, d), such that their dot product is zero: a*c + b*d = 0. Also, since they are direction vectors, we can assume they are unit vectors, so a² + b² = 1 and c² + d² = 1.Given that the first ray is incident at angle theta1, its direction vector before refraction is (sin(theta1), 0, -cos(theta1)), right? Because the angle with the normal (z-axis) is theta1, so the direction vector has a component along the z-axis of -cos(theta1) (since it's going downward) and a component along the x-axis of sin(theta1). Similarly, for the second ray, if it's incident at angle theta1', its direction vector is (sin(theta1'), 0, -cos(theta1')).The dot product of these two vectors is sin(theta1)sin(theta1') + cos(theta1)cos(theta1') = cos(theta1 - theta1'). For them to be perpendicular, this must be zero, so theta1 - theta1' = ±90°. But as before, theta1 and theta1' are both between 0° and 90°, so theta1 - theta1' = 90° would mean theta1 = theta1' + 90°, which would make theta1 greater than 90°, which is not possible. Similarly, theta1 - theta1' = -90° would mean theta1' = theta1 + 90°, which is also impossible.Therefore, this suggests that there is a contradiction, which can't be. Therefore, my assumption must be wrong.Wait, but the problem states that two mutually perpendicular rays are incident on the glass surface. So, perhaps they are not both incident from air, but that seems not the case. The problem says "fall from air onto a horizontal glass surface". So both rays are coming from air. Then, if their direction vectors are perpendicular, but in the same vertical plane, that seems impossible unless one of them is at an angle greater than 90°, which is not possible for incidence.Wait, maybe the problem is that the two rays are perpendicular before refraction, but after refraction, they are not necessarily perpendicular? Wait, the problem states "two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface". So the mutual perpendicularity is in air, before refraction. So the two incident rays are perpendicular to each other. Then, after refraction, we have to find the angle between the two refracted rays.Ah! That's probably the correct interpretation. The two incident rays are mutually perpendicular in air, and upon entering glass, their refracted rays are no longer perpendicular. We need to find the angle between the two refracted rays.Okay, that makes more sense. So the angle between the two incident rays is 90 degrees. Then, each ray is refracted into the glass, and we need to find the angle between the two refracted rays.Given that the first refracted ray has an angle beta = 25 degrees. So, for the first ray, we can find the angle of incidence using Snell's Law, then find the direction of the refracted ray. For the second ray, since it is perpendicular to the first incident ray, we can find its angle of incidence, apply Snell's Law to find its angle of refraction, then compute the angle between the two refracted rays.Wait, but how do we find the angle of incidence for the second ray, given that it's perpendicular to the first incident ray? Let's clarify this.Let me denote the first incident ray as having an angle of incidence theta1, which we found earlier to be arcsin(1.5 * sin(25°)) ≈ 39.3 degrees. Then, the second incident ray is perpendicular to the first incident ray. So, the angle between their directions is 90 degrees.So, similar to before, if the first incident ray has a direction vector (sin(theta1), 0, -cos(theta1)), then the second incident ray must have a direction vector that is perpendicular to this. Let's denote the direction vector of the second incident ray as (a, 0, b), such that (sin(theta1), 0, -cos(theta1)) • (a, 0, b) = 0. So:sin(theta1) * a + (-cos(theta1)) * b = 0Also, since the second incident ray is propagating in air, its direction vector must be a unit vector, so a² + b² = 1. Additionally, since it's incident on the glass surface, the z-component (b) should be negative (assuming the surface is at z=0 and the rays are coming from z>0).Wait, but if the first ray's direction vector is (sin(theta1), 0, -cos(theta1)), which points downward towards the surface, then the second ray's direction vector (a, 0, b) must also point downward, so b is negative.So, solving sin(theta1)*a - cos(theta1)*b = 0 and a² + b² = 1.Let me express a in terms of b from the first equation:a = (cos(theta1)/sin(theta1)) * bThen substitute into the second equation:( (cos(theta1)/sin(theta1)) * b )² + b² = 1=> (cos²(theta1)/sin²(theta1)) * b² + b² = 1=> b² [ cos²(theta1)/sin²(theta1) + 1 ] = 1Factor out b²:b² [ (cos²(theta1) + sin²(theta1))/sin²(theta1) ] = 1Since cos² + sin² = 1:b² [ 1 / sin²(theta1) ] = 1Therefore:b² = sin²(theta1)So, b = ± sin(theta1)But since the z-component is negative (as the ray is going downward), we take b = -sin(theta1)Wait, wait. Wait, no. Wait, b is the z-component. If the direction vector is (a, 0, b), and the ray is going downward, then b should be negative. But in the previous step, we have b = ± sin(theta1). Wait, let's see:From b² = sin²(theta1), we get b = ± sin(theta1). But since the direction is downward, we take b = -sin(theta1). Then, a = (cos(theta1)/sin(theta1)) * b = (cos(theta1)/sin(theta1)) * (-sin(theta1)) = -cos(theta1)So, the direction vector of the second incident ray is (-cos(theta1), 0, -sin(theta1))Wait, let's check the unit vector:a² + b² = (-cos(theta1))² + (-sin(theta1))² = cos²(theta1) + sin²(theta1) = 1. Good.And the dot product with the first incident direction vector:sin(theta1)*(-cos(theta1)) + (-cos(theta1))*(-sin(theta1)) = -sin(theta1)cos(theta1) + sin(theta1)cos(theta1) = 0. Perfect.So, the second incident ray's direction vector is (-cos(theta1), 0, -sin(theta1)). But what is its angle of incidence?The angle of incidence is measured with respect to the normal (z-axis). So, the angle of incidence theta1' is the angle between the direction vector and the normal. Since the direction vector is (-cos(theta1), 0, -sin(theta1)), the angle theta1' can be found by taking the arccos of the z-component divided by the magnitude (which is 1). But since the direction vector is pointing downward, the angle is measured from the positive z-axis to the direction vector. However, the z-component is -sin(theta1), but wait:Wait, no. Wait, the direction vector is (-cos(theta1), 0, -sin(theta1)). To find the angle with the normal (positive z-axis), we can use the dot product formula:cos(theta1') = direction_vector • (0, 0, 1) / |direction_vector| = (-sin(theta1)) / 1 = -sin(theta1)But theta1' is the angle between the direction vector and the normal (positive z-axis), which is in [0, 180°]. However, the angle of incidence is defined as the angle between the incident ray and the normal, which is always between 0° and 90°. Therefore, theta1' = 180° - arccos(-sin(theta1)).But let's compute this. Let's suppose theta1 is 39.3°, as we found earlier.So, cos(theta1') = -sin(theta1) ≈ -sin(39.3°) ≈ -0.6339Therefore, theta1' = arccos(-0.6339) ≈ 129.3°, but since angle of incidence is the acute angle between the ray and the normal, we take 180° - 129.3° ≈ 50.7°. Wait, that can't be. Wait, no, the angle of incidence is defined as the angle between the incident ray and the normal, measured in the incident medium (air). Since the direction vector is pointing downward, the angle of incidence is the angle between the ray and the normal (z-axis), which is measured from the normal to the ray. So, if the direction vector is (-cos(theta1), 0, -sin(theta1)), then the angle theta1' is the angle between the negative z-axis and the direction vector?Wait, perhaps a better approach is to realize that the angle of incidence is the angle between the ray and the normal. So, if the direction vector is (-cos(theta1), 0, -sin(theta1)), then the angle theta1' can be found by taking the angle whose tangent is the ratio of the horizontal component to the vertical component. The horizontal component is the x-component, which is -cos(theta1), and the vertical component is the z-component, which is -sin(theta1). So, tan(theta1') = |horizontal| / |vertical| = | -cos(theta1) | / | -sin(theta1) | = cos(theta1)/sin(theta1) = cot(theta1)Therefore, tan(theta1') = cot(theta1) = tan(90° - theta1)Therefore, theta1' = 90° - theta1Ah! So, the angle of incidence for the second ray is theta1' = 90° - theta1. So, if theta1 is 39.3°, theta1' is 90° - 39.3° = 50.7°. Therefore, the second ray has an angle of incidence of 50.7 degrees. Then, using Snell's Law, we can find its angle of refraction.Let me compute that. Using Snell's Law for the second ray: n1 * sin(theta1') = n2 * sin(theta2'), where theta2' is the angle of refraction. So:sin(theta2') = (n1 / n2) * sin(theta1') = (1 / 1.5) * sin(50.7°)First, compute sin(50.7°). Let's see, sin(50.7°) ≈ 0.774. So, 0.774 / 1.5 ≈ 0.516. Therefore, theta2' ≈ arcsin(0.516) ≈ 31 degrees.Therefore, the second refracted ray has an angle of 31 degrees with the normal.Now, the first refracted ray is at beta = 25 degrees from the normal, and the second refracted ray is at 31 degrees from the normal. Now, we need to find the angle between these two refracted rays.Since both refracted rays are in the same vertical plane (the plane of incidence), their direction vectors are in that plane. The angle between them can be found by subtracting their angles with respect to the normal. But wait, if both are on the same side of the normal, then the angle between them is theta2' - beta. But depending on their directions, it might be the sum. Wait, let me think.The first refracted ray is at 25 degrees below the normal (since it's going into the glass), and the second refracted ray is at 31 degrees below the normal. Since both are in the same vertical plane, the angle between them is the difference between their angles of refraction. But is it the difference or the sum?Wait, if both rays are on the same side of the normal, then the angle between them is the absolute difference between their angles. However, if they are on opposite sides, it's the sum. In this case, since both rays are refracted towards the normal (as they go from air to glass), they are both on the same side of the normal. Therefore, the angle between them is |31° - 25°| = 6°. Wait, but this seems too small.Wait, no. Wait, actually, the angle between the two refracted rays is the angle between their direction vectors. Since both are in the same vertical plane, with angles beta and theta2' with respect to the normal, which is the vertical. So, if the first refracted ray is at 25 degrees from the normal (vertical), and the second is at 31 degrees from the normal, then the angle between them is 25° + 31° = 56°? Wait, that doesn't make sense either.Wait, let's clarify. Let's imagine the normal is vertical. The first refracted ray is at 25 degrees to the normal, so it's going to the right and down at 25 degrees. The second refracted ray is at 31 degrees to the normal. But wait, how is the second refracted ray oriented? Since the second incident ray was coming from a different direction (perpendicular to the first incident ray), the direction of the second refracted ray might be on the opposite side of the normal?Wait, no. Both refracted rays are in the same vertical plane. The first incident ray was coming from, say, the right side, making an angle theta1 with the normal. The second incident ray was coming from a direction perpendicular to the first, so in the vertical plane, but in which direction?Wait, the direction vector of the second incident ray was (-cos(theta1), 0, -sin(theta1)). So, in the x-z plane, the x-component is negative, meaning it's coming from the left side. So, the second incident ray is coming from the left, at an angle theta1' = 90° - theta1 with the normal.Therefore, when it refracts into the glass, it will bend towards the normal. So, its refracted direction will be on the left side of the normal, at an angle theta2' = arcsin( (1/1.5) sin(theta1') ) ≈ 31 degrees.Meanwhile, the first refracted ray is on the right side of the normal at 25 degrees.Therefore, the two refracted rays are on opposite sides of the normal, each at 25 degrees and 31 degrees from the normal. Therefore, the angle between them is 25° + 31° = 56°.Yes, that makes sense. Because if one is refracted to the right at 25°, and the other to the left at 31°, the total angle between them is the sum of their angles from the normal. Therefore, the angle between the two refracted rays is 25° + 31° = 56°.Therefore, the answer should be 56 degrees.Wait, but let me confirm this with vector calculations.Let's define the refracted rays' direction vectors.For the first refracted ray, the angle with the normal is beta = 25°, so its direction vector in glass is (sin(beta), 0, -cos(beta)). Wait, no, since it's moving downward into the glass, the z-component is negative. Wait, actually, in the glass, the direction would still be downward, so the z-component is negative. But in terms of angles, if beta is the angle from the normal (z-axis), then the direction vector is (sin(beta), 0, -cos(beta)).Similarly, for the second refracted ray, the angle with the normal is theta2' = 31°, but since it's on the opposite side of the normal (left side), the x-component would be negative. Therefore, its direction vector is (-sin(theta2'), 0, -cos(theta2')).Now, to find the angle between these two vectors, we can compute the dot product.Dot product = [sin(beta) * (-sin(theta2'))] + [0 * 0] + [ -cos(beta) * (-cos(theta2')) ]= -sin(beta)sin(theta2') + cos(beta)cos(theta2')= cos(beta + theta2') [Using cosine addition formula: cos(A + B) = cosA cosB - sinA sinB]Wait, wait: cos(beta - theta2') = cos(beta)cos(theta2') + sin(beta)sin(theta2'). But here we have cos(beta)cos(theta2') - sin(beta)sin(theta2') which is cos(beta + theta2'). Therefore, the dot product is cos(beta + theta2').But the angle between the two vectors is phi, such that cos(phi) = dot product divided by the product of their magnitudes. Since both vectors are unit vectors (assuming the glass has the same speed, but actually, in reality, the direction vectors in the medium would have a magnitude related to the wavelength, but for the purpose of angle calculation, direction vectors are considered unit vectors). Therefore, cos(phi) = cos(beta + theta2'). Therefore, phi = beta + theta2'.Wait, but beta is 25°, theta2' is 31°, so phi = 56°, which matches our previous conclusion. Therefore, the angle between the refracted rays is 56 degrees.Therefore, the answer is 56 degrees, which in boxed form is boxed{56^circ}.Final Answerboxed{56^circ}</think>Given two mutually perpendicular rays propagating in the same vertical plane and falling onto a horizontal glass surface with a refractive index ( n = 1.5 ), we need to determine the angle between the refracted rays. The angle of refraction for the first ray is given as ( beta = 25^circ ).1. Finding the angle of incidence for the first ray: Using Snell's Law, ( n_1 sin(theta_1) = n_2 sin(beta) ): [ sin(theta_1) = frac{n_2}{n_1} sin(beta) = 1.5 sin(25^circ) ] Calculating ( sin(25^circ) approx 0.4226 ): [ sin(theta_1) approx 1.5 times 0.4226 approx 0.6339 implies theta_1 approx arcsin(0.6339) approx 39.3^circ ]2. Determining the angle of incidence for the second ray: Since the rays are mutually perpendicular, the angle of incidence for the second ray ( theta_1' ) is ( 90^circ - theta_1 ): [ theta_1' = 90^circ - 39.3^circ approx 50.7^circ ]3. Finding the angle of refraction for the second ray: Using Snell's Law again: [ sin(beta') = frac{n_1}{n_2} sin(theta_1') = frac{1}{1.5} sin(50.7^circ) ] Calculating ( sin(50.7^circ) approx 0.774 ): [ sin(beta') approx frac{0.774}{1.5} approx 0.516 implies beta' approx arcsin(0.516) approx 31^circ ]4. Calculating the angle between the refracted rays: The angle between the two refracted rays is the sum of their angles of refraction relative to the normal since they are on opposite sides of the normal: [ text{Angle between refracted rays} = 25^circ + 31^circ = 56^circ ]Thus, the angle between the refracted rays is (boxed{56^circ}).]

🔑:Answer: 56^{circ}.Solution. The law of refraction for the first ray frac{sin alpha}{sin 25^{circ}}=1.5. Therefore, the angle of incidence of the first ray alpha=39.34^{circ}. The angle of incidence of the second ray beta=90^{circ}-39.34^{circ}= 50.66^{circ}. The law of refraction for the second ray frac{sin 50.66^{circ}}{sin gamma}=1.5. The angle of refraction of the second ray gamma=31^{circ}. The angle between the refracted rays: beta+gamma=56^{circ}.

🧠:[<think>Okay, so I need to find the complex number z that satisfies the equation z multiplied by i equals 3 minus 4i. Let me think about how to approach this. First, I remember that when dealing with complex numbers, multiplying by i is equivalent to rotating the complex number by 90 degrees in the complex plane. But maybe that's a bit abstract for solving this equation directly. Let me try to recall algebraic methods instead.If I have z * i = 3 - 4i, then to solve for z, I should divide both sides by i. But dividing complex numbers can be a bit tricky. Alternatively, I can represent z as a complex number in the form a + bi, where a and b are real numbers, and then substitute into the equation to solve for a and b. That might be more straightforward.So let's let z = a + bi. Then the equation becomes (a + bi) * i = 3 - 4i. Let's compute the left side:Multiplying (a + bi) by i: (a * i) + (bi * i) = ai + b(i^2). Since i^2 = -1, this simplifies to ai + b(-1) = -b + ai.So, the left side is -b + ai. The right side is 3 - 4i. Now, we can equate the real and imaginary parts on both sides. The real part of the left side is -b, and the imaginary part is a. The real part of the right side is 3, and the imaginary part is -4. Therefore, setting the real parts equal:-b = 3 => b = -3And setting the imaginary parts equal:a = -4So, a is -4 and b is -3. Therefore, z = a + bi = -4 - 3i. Let me check the options. Option D is -4 -3i, so that should be the answer.Wait, but let me verify this by another method to be sure. If z = -4 -3i, then multiplying by i:z * i = (-4 -3i) * i = -4i -3i^2 = -4i -3(-1) = -4i + 3 = 3 -4i. Yep, that matches the equation given. So that's correct.Alternatively, another method is to divide both sides by i. To divide complex numbers, you can multiply numerator and denominator by the complex conjugate of the denominator. Wait, but in this case, the denominator is i. So maybe it's easier to note that 1/i = -i, since i * -i = 1.So starting from z * i = 3 -4i, then z = (3 -4i)/i. Multiply numerator and denominator by i: (3 -4i)*i / (i*i). The denominator becomes i^2 = -1. The numerator becomes 3i -4i^2 = 3i -4(-1) = 3i +4. So z = (4 +3i)/(-1) = -4 -3i. Which is the same result. So that confirms it.Therefore, the answer is D: -4 -3i.Final Answerboxed{D}</think>Given the complex number ( z ) that satisfies the equation ( z cdot i = 3 - 4i ), we need to find ( z ).First, we represent ( z ) as ( a + bi ), where ( a ) and ( b ) are real numbers. Substituting this into the equation, we get:[(a + bi) cdot i = 3 - 4i]Expanding the left side:[ai + bi^2 = ai - b]Equating the real and imaginary parts from both sides:- Real part: (-b = 3 implies b = -3)- Imaginary part: (a = -4)Thus, ( z = a + bi = -4 - 3i ).To verify, we multiply ( z ) by ( i ):[(-4 - 3i) cdot i = -4i - 3i^2 = -4i + 3 = 3 - 4i]This confirms the solution is correct. Another verification method involves dividing both sides by ( i ):[z = frac{3 - 4i}{i} cdot frac{-i}{-i} = frac{(3 - 4i)(-i)}{1} = -3i + 4i^2 = -3i - 4 = -4 - 3i]Thus, the answer is (boxed{D}).,<think>Okay, so I need to find the complex number z that satisfies the equation z multiplied by i equals 3 minus 4i. Let's see... Hmm, complex numbers can sometimes be tricky, but maybe if I approach it step by step, it'll make more sense.First, let's recall that complex numbers are in the form a + bi, where a and b are real numbers, and i is the imaginary unit with the property that i² = -1. So, if z is a complex number, I can write z as a + bi. Then, when I multiply z by i, according to the problem, that product equals 3 - 4i. So let's write that out. If z = a + bi, then z * i = (a + bi) * i. Let me compute that. Distributing the multiplication, that would be a*i + bi*i. Which simplifies to ai + b(i²). Since i² is -1, that becomes ai + b(-1) = ai - b. So, z * i = -b + ai. But according to the equation, z * i = 3 - 4i. So, setting these equal, we have:- b + ai = 3 - 4i.Now, for two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. So, the real part of the left side is -b, and the imaginary part is a. On the right side, the real part is 3, and the imaginary part is -4. Therefore, we can set up the equations:1. Real parts: -b = 32. Imaginary parts: a = -4Solving equation 1: -b = 3 => b = -3.Solving equation 2: a = -4.Therefore, z = a + bi = -4 + (-3)i = -4 - 3i. Let me check the options. Option D is -4 - 3i. So that should be the answer. Wait, but let me verify again to make sure I didn't make any mistakes.Alternatively, maybe there's another way to solve for z. If z * i = 3 - 4i, then to solve for z, I can divide both sides by i. But how does division work with complex numbers? Oh, right, you can multiply the numerator and denominator by the complex conjugate of the denominator. But since the denominator here is just i, which is a purely imaginary number, perhaps it's simpler. Let's try that method as a check.Starting with z * i = 3 - 4i. To solve for z, divide both sides by i:z = (3 - 4i)/i.Hmm, dividing by i. Let's see. Multiply numerator and denominator by i to rationalize the denominator:z = [(3 - 4i)/i] * [i/i] = [(3 - 4i)*i]/(i*i).Compute denominator: i*i = i² = -1.Compute numerator: (3 - 4i)*i = 3i - 4i². Since i² = -1, that becomes 3i -4*(-1) = 3i +4.So numerator is 4 + 3i, denominator is -1. Therefore:z = (4 + 3i)/(-1) = -4 -3i.Same result as before. So z = -4 -3i, which is option D. That seems consistent. But let me double-check to make absolutely sure.Alternatively, maybe plug the options back into the original equation to see which one works. Let's try that with each option:Option A: z = 3 -4i. Then z * i = (3 -4i)*i = 3i -4i² = 3i -4(-1) = 3i +4 = 4 + 3i. But the problem states z*i = 3 -4i. So 4 + 3i ≠ 3 -4i. So A is wrong.Option B: z =4 +3i. Then z*i = (4 +3i)*i =4i +3i² =4i +3(-1)= -3 +4i. Comparing to 3 -4i, not the same. So B is wrong.Option C: z =-3 +4i. Then z*i= (-3 +4i)*i= -3i +4i²= -3i +4(-1)= -4 -3i. But the right side is 3 -4i, so that's not matching. So C is wrong.Option D: z= -4 -3i. Then z*i= (-4 -3i)*i= -4i -3i²= -4i -3(-1)= -4i +3= 3 -4i. Which is exactly the equation given. So D is correct.Therefore, the answer is D. Wait, but just to make sure, when I multiplied -4 -3i by i, let's redo that:(-4 -3i) * i = (-4)*i + (-3i)*i = -4i -3i². Since i² = -1, so -3i² = -3*(-1) = 3. So total is -4i +3, which is 3 -4i. Perfect, matches the equation given. So yes, D is correct.Therefore, the correct answer is D: -4 -3i.Final Answerboxed{D}</think>Given the complex number ( z ) that satisfies the equation ( z cdot i = 3 - 4i ), we need to find ( z ).First, we express ( z ) as ( a + bi ) where ( a ) and ( b ) are real numbers. Then, we compute ( z cdot i ):[z cdot i = (a + bi) cdot i = ai + bi^2 = ai - b]Given ( z cdot i = 3 - 4i ), we equate the real and imaginary parts:1. Real part: (-b = 3)2. Imaginary part: (a = -4)Solving these equations, we find ( b = -3 ) and ( a = -4 ). Therefore, ( z = -4 - 3i ).To verify, we can divide both sides of the equation ( z cdot i = 3 - 4i ) by ( i ):[z = frac{3 - 4i}{i}]Multiplying the numerator and denominator by ( i ):[z = frac{(3 - 4i) cdot i}{i cdot i} = frac{3i - 4i^2}{i^2} = frac{3i + 4}{-1} = -4 - 3i]Checking the options, substituting each into the original equation confirms that option D is correct:- Option D: ( z = -4 - 3i )- ( z cdot i = (-4 - 3i) cdot i = -4i - 3i^2 = -4i + 3 = 3 - 4i )Thus, the correct answer is (boxed{D}).]

🔑:From z cdot i = 3 - 4i, we get z = frac{3 - 4i}{i} = frac{(3 - 4i)(-i)}{-i^2} = -4 - 3i.So the answer is: boxed{text{D}}.We transform the given equation and then simplify it using the algebraic form of complex numbers multiplication and division to get the answer. This problem tests the understanding of complex number's algebraic form multiplication and division, which is a basic concept.

🧠:[<think>Okay, let's see. The problem says that O is the circumcenter of triangle ABC, and the vector equation OA + OB = OC holds. We need to find the measure of angle ACB. Hmm, circumcenter means O is equidistant from all three vertices, right? So OA = OB = OC. That's because the circumcenter is the intersection of the perpendicular bisectors of the sides. But the vector equation here is interesting. Vector OA plus vector OB equals vector OC. Let me recall some vector properties. If vectors add up like that, maybe we can interpret it geometrically. Since O is the origin of these vectors (since they're all starting from O), the sum OA + OB would be the diagonal of the parallelogram formed by OA and OB. But that diagonal should be equal to OC. So, if OA and OB are sides of a parallelogram, then the diagonal is OC. But since OA, OB, and OC are all radii of the circumcircle, their magnitudes are equal. So OA, OB, and OC have the same length. Wait, so if OA and OB are vectors of the same length, and their sum is OC, which is also the same length, that must mean something about the angle between OA and OB. Let me think. If we have two vectors of equal length, say |a| = |b| = r, then |a + b| = |c| = r. Let's compute the magnitude squared. |a + b|² = |c|² = r². But |a + b|² = |a|² + |b|² + 2a·b = r² + r² + 2a·b = 2r² + 2a·b. Therefore, 2r² + 2a·b = r² ⇒ 2a·b = -r² ⇒ a·b = -r²/2. But a·b = |a||b|cosθ = r² cosθ, where θ is the angle between vectors a and b. Therefore, r² cosθ = -r²/2 ⇒ cosθ = -1/2 ⇒ θ = 120 degrees. So the angle between OA and OB is 120 degrees. But OA and OB are vectors from the circumcenter to vertices A and B. The angle between OA and OB is the central angle corresponding to arc AB. In a circle, the central angle is twice the inscribed angle subtended by the same arc. So the inscribed angle at point C (which is angle ACB) would be half of 120 degrees, which is 60 degrees. Wait, is that correct? Let me verify. If the central angle for arc AB is 120 degrees, then the inscribed angle subtended by arc AB would be 120/2 = 60 degrees. But angle ACB is the angle at point C, and depending on the position of point C, the inscribed angle could be different. Wait, no. The inscribed angle depends on the arc opposite to it. So angle ACB is subtended by arc AB, so if the central angle for arc AB is 120 degrees, then angle ACB is indeed 60 degrees. But let me make sure. Let me sketch triangle ABC with circumcenter O. If OA + OB = OC, then based on the vector addition, point C is such that OC is the diagonal of the parallelogram formed by OA and OB. Since OA and OB are radii, and their sum is also a radius, the parallelogram must be a rhombus? Wait, no. Because OA and OB have the same length, and their sum OC also has the same length. That would mean that the rhombus formed by OA and OB has diagonals equal to OC. Wait, maybe not. Let's think again.If OA and OB are vectors from O, then OA + OB is the vector to point C. But since OA, OB, and OC are all radii, the triangle is such that C is located at the tip of the vector OA + OB, but OA + OB must have the same length as OA and OB. So, in the complex plane, if we represent points A, B, C as complex numbers of magnitude r (the radius), then A + B = C. Then |A + B| = |C| = r. Squaring both sides, |A + B|² = |C|² = r². Then, expanding |A + B|², we get |A|² + |B|² + 2 Re(A overline{B}) = r² + r² + 2 Re(A overline{B}) = 2r² + 2 Re(A overline{B}) = r². Therefore, 2 Re(A overline{B}) = -r². Since Re(A overline{B}) is the dot product of vectors A and B (if we consider them as vectors in 2D), which we already found to be -r²/2. So the angle between OA and OB is 120 degrees, as before. Therefore, the central angle for arc AB is 120 degrees. Then the inscribed angle at C for arc AB is 60 degrees. Therefore, angle ACB is 60 degrees. So the answer is 60 degrees, which is π/3 radians. So in boxed form, boxed{60^circ}.Wait, but let me check if there's another possibility. Suppose the triangle is oriented differently. For example, could angle ACB be 120 degrees? Let me see. If the central angle is 120 degrees, then the inscribed angle is 60 degrees. But if the central angle is 60 degrees, the inscribed angle would be 30 degrees. Wait, but we already found the central angle is 120 degrees, so the inscribed angle must be 60. So that should be correct. Alternatively, maybe there's another way to approach this problem. Let's use coordinate geometry. Let me place the circumcenter O at the origin for simplicity. Then, vectors OA, OB, OC are just position vectors of points A, B, C. Given OA + OB = OC. So, if OA is vector (a, b) and OB is vector (c, d), then OC is (a + c, b + d). But since OA, OB, and OC all have the same magnitude (since it's the circumradius), so |OA|² = a² + b² = |OB|² = c² + d² = |OC|² = (a + c)² + (b + d)². So, expanding |OC|²: (a + c)^2 + (b + d)^2 = a² + 2ac + c² + b² + 2bd + d² = (a² + b²) + (c² + d²) + 2(ac + bd) = |OA|² + |OB|² + 2(ac + bd). But since |OA| = |OB| = |OC|, let's denote |OA| = |OB| = |OC| = r. Therefore, |OC|² = r² = r² + r² + 2(ac + bd) ⇒ r² = 2r² + 2(ac + bd) ⇒ -r² = 2(ac + bd) ⇒ ac + bd = -r² / 2. But ac + bd is the dot product of vectors OA and OB. So OA · OB = -r² / 2. As before, the dot product is |OA||OB|cosθ = r² cosθ. So r² cosθ = -r² / 2 ⇒ cosθ = -1/2 ⇒ θ = 120 degrees. So again, the central angle is 120 degrees, leading to the inscribed angle of 60 degrees. Therefore, angle ACB is 60 degrees. Wait, another approach: in triangle ABC, since O is the circumcenter, OA=OB=OC=R. Then, given OA + OB = OC. Let me try to use coordinate geometry. Let me place point O at the origin. Let me assign coordinates such that point A is at (R, 0). Then, point B can be at some coordinate (R cos α, R sin α). Then, OA vector is (R, 0), OB vector is (R cos α, R sin α). Then, OC vector is OA + OB = (R + R cos α, 0 + R sin α). But OC must have magnitude R. Therefore:|OC|² = (R + R cos α)^2 + (R sin α)^2 = R²(1 + 2 cos α + cos² α + sin² α) = R²(2 + 2 cos α) because cos² α + sin² α =1. So |OC|² = 2 R² (1 + cos α). But |OC|² must equal R². Therefore:2 R² (1 + cos α) = R² ⇒ 2(1 + cos α) =1 ⇒ 1 + cos α = 1/2 ⇒ cos α = -1/2 ⇒ α = 120 degrees or 240 degrees. But since α is the angle between OA and OB, which is 120 degrees. So the angle between OA and OB is 120 degrees, so central angle AOB is 120 degrees. Then, angle ACB, which is the inscribed angle subtended by arc AB, is half of 120 degrees, which is 60 degrees. So all approaches lead to angle ACB being 60 degrees. Therefore, the answer should be 60 degrees. But to be thorough, let me think if there's a case where angle ACB could be 120 degrees. Wait, if the triangle is obtuse, but in that case, the circumradius is still positive. Wait, no. The inscribed angle can't be more than 180 degrees, but in this case, since the central angle is 120, the inscribed angle is 60. If the central angle was 240 degrees, then the inscribed angle would be 120 degrees, but in that case, the position of point C would be different. Wait, but in our problem, OC is given as OA + OB. So when we calculated alpha as 120 degrees, that's the angle between OA and OB. If alpha is 240 degrees, then cos alpha = -1/2 as well, but 240 degrees is equivalent to -120 degrees. But in terms of the central angle, 240 degrees would mean that point B is on the other side, but since OA and OB are vectors, adding them would result in OC being in a different position. However, in our coordinate system, we fixed point A at (R, 0), and point B at (R cos alpha, R sin alpha). If alpha is 240 degrees, then point B would be at (R cos 240°, R sin 240°) = ( -R/2, -R√3/2 ). Then OA + OB would be (R - R/2, 0 - R√3/2 ) = (R/2, -R√3/2 ). The magnitude of this vector is sqrt( (R/2)^2 + ( -R√3/2 )^2 ) = sqrt( R²/4 + 3 R²/4 ) = sqrt(R²) = R. So OC is indeed a vector of magnitude R. But in this case, where is point C? At (R/2, -R√3/2 ). So angle ACB would be... Hmm, need to compute angle at point C between points A and B. Wait, but in this configuration, points A is (R,0), B is (-R/2, -R√3/2 ), and C is (R/2, -R√3/2 ). Let me compute angle ACB. First, find coordinates:A: (R, 0)B: (-R/2, -R√3/2 )C: (R/2, -R√3/2 )Compute vectors CA and CB.Vector CA = A - C = (R - R/2, 0 - (-R√3/2 )) = (R/2, R√3/2 )Vector CB = B - C = (-R/2 - R/2, -R√3/2 - (-R√3/2 )) = (-R, 0 )So angle at C is the angle between vectors CA and CB. Vector CA is (R/2, R√3/2 ), vector CB is (-R, 0 ). The angle between them can be found using the dot product:cosθ = (CA · CB) / (|CA| |CB| )CA · CB = (R/2)(-R) + (R√3/2)(0) = -R²/2|CA| = sqrt( (R/2)^2 + (R√3/2 )^2 ) = sqrt( R²/4 + 3 R²/4 ) = sqrt(R²) = R|CB| = sqrt( (-R)^2 + 0^2 ) = RTherefore, cosθ = (-R²/2 ) / (R * R ) = -1/2 ⇒ θ = 120 degreesWait, that's different! So when alpha is 240 degrees, angle ACB is 120 degrees. But in the problem statement, we have OA + OB = OC. So in this coordinate system, OC is (R/2, -R√3/2 ). So point C is (R/2, -R√3/2 ). Then angle ACB is 120 degrees. But earlier when alpha was 120 degrees, angle ACB was 60 degrees. So why the discrepancy?Wait, this suggests that there are two possible configurations: one where angle between OA and OB is 120 degrees, resulting in angle ACB being 60 degrees, and another where angle between OA and OB is 240 degrees (which is equivalent to -120 degrees), resulting in angle ACB being 120 degrees. But how does that fit with the problem statement?Wait, but the problem statement just gives the vector equation OA + OB = OC. In our coordinate system, when alpha is 120 degrees, point C is (R + R cos 120°, R sin 120° ) = (R - R/2, R*(√3/2)) = (R/2, R√3/2 ). Then angle ACB in that case would be different. Let me compute angle ACB in that case.So points:A: (R,0)B: (R cos 120°, R sin 120° ) = (-R/2, R√3/2 )C: (R/2, R√3/2 )Vectors CA and CB:CA = A - C = (R - R/2, 0 - R√3/2 ) = (R/2, -R√3/2 )CB = B - C = (-R/2 - R/2, R√3/2 - R√3/2 ) = (-R, 0 )So vectors CA = (R/2, -R√3/2 ) and CB = (-R, 0 )Dot product: (R/2)(-R) + (-R√3/2)(0) = -R²/2|CA| = sqrt( (R/2)^2 + ( - R√3/2 )^2 ) = sqrt( R²/4 + 3 R²/4 ) = R|CB| = sqrt( (-R)^2 + 0^2 ) = RTherefore, cosθ = (-R²/2)/(R*R) = -1/2 ⇒ θ = 120 degrees again. Wait, so in both cases, angle ACB is 120 degrees? Wait, this contradicts my previous conclusion. What's going on here?Wait, no. Wait, in both configurations where alpha is 120 degrees and 240 degrees, angle ACB is 120 degrees? But that can't be. Wait, no, in the first case where alpha was 120 degrees, but in that coordinate system, point C was (R/2, R√3/2 ), so in that case, points A, B, C are arranged such that angle at C is 120 degrees? Wait, but when alpha is 120 degrees, the central angle AOB is 120 degrees, so the inscribed angle ACB should be 60 degrees. But according to the coordinate calculation, it's 120 degrees. So which is it?Wait, perhaps I made an error in the coordinate calculation. Let me check again. First configuration: O at origin, A at (R,0), B at (R cos 120°, R sin 120° ) = (-R/2, R√3/2 ). Then OC = OA + OB = (R,0) + (-R/2, R√3/2 ) = (R - R/2, 0 + R√3/2 ) = (R/2, R√3/2 ). So point C is (R/2, R√3/2 ).Then vectors:CA = A - C = (R - R/2, 0 - R√3/2 ) = (R/2, -R√3/2 )CB = B - C = (-R/2 - R/2, R√3/2 - R√3/2 ) = (-R, 0 )So vectors CA and CB are (R/2, -R√3/2 ) and (-R, 0 ). The angle between these vectors is calculated as:cosθ = ( (R/2)(-R) + (-R√3/2)(0) ) / ( |CA| |CB| ) = (-R²/2) / (R * R ) = -1/2 ⇒ θ = 120 degrees.But according to the central angle, angle AOB is 120 degrees, so inscribed angle ACB should be 60 degrees. But here, it's 120 degrees. What's the problem?Wait, perhaps the arc that angle ACB subtends. Wait, angle ACB is an inscribed angle. Which arc does it subtend? It subtends arc AB. If central angle AOB is 120 degrees, then inscribed angle ACB should be half of that, 60 degrees. But according to coordinates, angle ACB is 120 degrees. Contradiction. Therefore, there's a mistake in my reasoning. Wait, maybe the position of point C is different. Wait, in the first scenario, when OA + OB = OC, point C is not on the circumcircle? But in the problem statement, O is the circumcenter, so OA=OB=OC. So OC must be a radius. In our calculation, OC has magnitude R, so it is on the circumcircle. But then why is angle ACB 120 degrees instead of 60?Wait, maybe I confused which arc the angle is subtended by. Let me recall that in a circle, the inscribed angle is equal to half the measure of its subtended arc. If angle ACB is at point C, then it subtends arc AB. If the central angle for arc AB is 120 degrees, then angle ACB is 60 degrees. However, in our coordinate calculation, we get angle ACB as 120 degrees. So why the discrepancy?Wait, perhaps because point C is not located on the same arc as before. Depending on the position of point C, the inscribed angle can be either θ or 180 - θ. Because if the central angle is 120 degrees, then the inscribed angle can be 60 degrees or 120 degrees, depending on which arc we take. Wait, the central angle is 120 degrees, so the corresponding inscribed angle could be 60 degrees if the points are on the same side, or 120 degrees if the point is on the opposite arc. Wait, but in the problem, we have triangle ABC with circumcenter O. So points A, B, C are all on the circumference. Then angle ACB is the inscribed angle subtended by arc AB. If the central angle for arc AB is 120 degrees, then angle ACB is 60 degrees. However, if angle ACB is on the opposite side of the circle, then it would subtend the other arc AB, which is 360 - 120 = 240 degrees, leading to an inscribed angle of 120 degrees. Ah! So depending on whether point C is on the minor arc AB or the major arc AB, the inscribed angle ACB will be 60 degrees or 120 degrees. So in our problem, since OA + OB = OC, where is point C located?In our coordinate system, when we added vectors OA and OB, we got point C at (R/2, R√3/2 ), which is above the x-axis. Points A and B are at (R,0) and (-R/2, R√3/2 ). So point C is also in the upper half-plane. In this configuration, the arc AB that doesn't contain C is 120 degrees, and the arc AB that contains C is 240 degrees. Wait, but the central angle for arc AB is 120 degrees. Wait, no. The central angle between OA and OB is 120 degrees. Therefore, the arc AB that corresponds to the central angle of 120 degrees is the minor arc AB. The major arc AB would be 240 degrees. Therefore, if point C is on the circumcircle, angle ACB subtends arc AB. If C is on the opposite side from the minor arc AB, then angle ACB is half the measure of the major arc AB, which is 240 / 2 = 120 degrees. If C is on the same side as the minor arc AB, then angle ACB is 60 degrees. In our coordinate calculation, point C is at (R/2, R√3/2 ), which is above the x-axis. Points A and B are (R,0) and (-R/2, R√3/2 ). So the minor arc AB is 120 degrees, and the major arc AB is 240 degrees. Point C is located such that when you go from A to B along the minor arc, you pass near point C? Wait, maybe not. Wait, point C is at (R/2, R√3/2 ). Let me check the positions:- Point A: (R,0)- Point B: (-R/2, R√3/2 )- Point C: (R/2, R√3/2 )So all three points are in the upper half-plane except point A is on the x-axis. So arc AB: moving from A (R,0) to B (-R/2, R√3/2 ) along the circumcircle, passing through the upper half would be the minor arc AB of 120 degrees, and the major arc AB would be 240 degrees going the other way. However, point C is also on the upper half. Wait, but when we connect points A, B, C, where is C? It's on the upper half as well. So does angle ACB subtend the minor arc AB or the major arc AB?Wait, in triangle ABC, angle at C subtends arc AB. If C is not on the arc AB, then the angle at C is half the measure of the arc AB that does not contain C. So if C is on the opposite arc, then the angle is half the measure of the arc AB that is opposite. Wait, this is confusing. Let me recall: in a circle, the inscribed angle theorem states that an angle subtended by an arc at the center is twice the angle subtended at the circumference. So if we have two points A and B on the circle, and a third point C on the circumference, then angle ACB is equal to half the measure of the arc AB that is not containing C. So if C is on the major arc AB, then angle ACB is half the measure of the minor arc AB, and vice versa. Wait, no. Wait, actually, the inscribed angle is half the measure of the arc that it subtends. So angle ACB subtends arc AB, regardless of where C is. Wait, no. If C is on the major arc AB, then angle ACB subtends the minor arc AB, and if C is on the minor arc AB, then angle ACB subtends the major arc AB. Wait, no, maybe it's the other way around. Let me look it up in my mind. The inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. The intercepted arc is the arc that lies in the interior of the angle. So if points A and B are fixed, and C is a point on the circumference, then angle ACB intercepts arc AB that is opposite to C. Wait, actually, the intercepted arc is the arc that is not containing the point C. So if C is on the major arc AB, then the intercepted arc is the minor arc AB, and the angle is half of that. If C is on the minor arc AB, then the intercepted arc is the major arc AB, and the angle is half of that. Wait, that can't be. Because if C is on the major arc AB, then angle ACB should be smaller. Wait, let me think of a semicircle. If AB is a diameter, then any point C on the circle forms a right angle, 90 degrees. The intercepted arc is 180 degrees, so angle is half, 90 degrees. If C is near A, then angle ACB is approaching 0 degrees, and the intercepted arc is approaching 0 degrees. Wait, so maybe the intercepted arc is the arc that is spanned by the angle. Alternatively, the intercepted arc for angle ACB is the arc AB that does not contain point C. Let me verify. Suppose the circle is divided into arc AB (let's say 120 degrees) and arc BA (240 degrees). If point C is on the arc AB (120 degrees), then angle ACB would subtend the opposite arc BA (240 degrees), so angle ACB would be 120 degrees. If point C is on the arc BA (240 degrees), then angle ACB would subtend arc AB (120 degrees), so angle ACB would be 60 degrees. Therefore, the inscribed angle is half the measure of the intercepted arc, which is the arc opposite to the angle. So if point C is on the opposite arc from AB, then angle ACB is half the measure of arc AB. If point C is on the same arc as AB, then angle ACB is half the measure of the opposite arc. Wait, this is getting confusing. Let's look at our specific case. Central angle AOB is 120 degrees. Then, the minor arc AB is 120 degrees, and the major arc AB is 240 degrees. If point C is not on the minor arc AB, then angle ACB subtends the minor arc AB and is 60 degrees. If point C is on the minor arc AB, then angle ACB subtends the major arc AB and is 120 degrees. In our problem, where is point C located? From the vector addition OA + OB = OC, we have point C determined by that vector sum. In our coordinate system, when OA is (R,0) and OB is at 120 degrees, point C is at (R/2, R√3/2 ), which is above the x-axis. So in this case, is point C on the minor arc AB or the major arc AB? Points A: (R,0), B: (-R/2, R√3/2 ), C: (R/2, R√3/2 ). The minor arc AB is 120 degrees. The major arc AB is 240 degrees. Point C is located at (R/2, R√3/2 ), which is between points A and another point, but not on the arc AB. Wait, actually, point C is outside the arc AB. Wait, no. To see if point C is on the minor or major arc AB, let's check the positions. The minor arc AB is the shorter path from A to B, which is 120 degrees. Starting at A (R,0), going counterclockwise 120 degrees reaches B (-R/2, R√3/2 ). Point C is at (R/2, R√3/2 ), which is 60 degrees counterclockwise from A. So point C is actually on the minor arc AB? Wait, no. Because from A (R,0), moving 60 degrees counterclockwise would reach (R cos 60°, R sin 60° ) = (R/2, R√3/2 ), which is point C. But the minor arc AB is 120 degrees from A to B. So point C is halfway along the minor arc AB. Therefore, point C is on the minor arc AB. Therefore, angle ACB subtends the major arc AB, which is 240 degrees, so angle ACB is 240 / 2 = 120 degrees. But earlier, we thought that since central angle AOB is 120 degrees, angle ACB is 60 degrees. But in reality, since point C is on the minor arc AB, angle ACB subtends the major arc AB, hence 240 / 2 = 120 degrees. Therefore, angle ACB is 120 degrees. But this contradicts our previous conclusion. So which one is correct? Wait, perhaps I need to check the position of point C again. If OA + OB = OC, then in our coordinate system, point C is at (R/2, R√3/2 ). Let's check if that point is on the circumcircle. Yes, because its distance from O is sqrt( (R/2)^2 + (R√3/2 )^2 ) = sqrt( R²/4 + 3 R²/4 ) = sqrt(R²) = R. So it is on the circumcircle. Now, considering triangle ABC with points A at (R,0), B at (-R/2, R√3/2 ), and C at (R/2, R√3/2 ). Let's compute angle ACB using coordinates. Coordinates:A: (R, 0)B: (-R/2, R√3/2 )C: (R/2, R√3/2 )Vectors:CA: A - C = (R - R/2, 0 - R√3/2 ) = (R/2, -R√3/2 )CB: B - C = (-R/2 - R/2, R√3/2 - R√3/2 ) = (-R, 0 )So vectors CA = (R/2, -R√3/2 ) and CB = (-R, 0 ). The angle between these vectors is calculated as follows:The dot product: (R/2)(-R) + (-R√3/2)(0) = -R²/2The magnitudes: |CA| = sqrt( (R/2)^2 + (-R√3/2 )^2 ) = sqrt( R²/4 + 3R²/4 ) = R|CB| = sqrt( (-R)^2 + 0^2 ) = RTherefore, cosθ = (-R²/2)/(R*R) = -1/2 ⇒ θ = 120 degrees. So according to the coordinate calculation, angle ACB is 120 degrees. But according to the earlier reasoning using central angles, it should be 60 degrees if point C is on the major arc AB. But in this case, point C is on the minor arc AB, so angle ACB is 120 degrees. Therefore, the correct answer is 120 degrees? But wait, the problem didn't specify where point C is located. It just says O is the circumcenter and OA + OB = OC. Wait, this is conflicting with the earlier conclusion. So now I'm confused. Which is correct?Wait, let's analyze again. The problem states that O is the circumcenter of triangle ABC, and OA + OB = OC. Then we need to find angle ACB. In our coordinate system, depending on the direction of the angle between OA and OB (120 or 240 degrees), we can get angle ACB as 60 or 120 degrees. But when we do the vector addition OA + OB = OC, in both cases, angle ACB comes out as 120 degrees. Wait, no. In the first case, when alpha was 120 degrees, angle ACB was 120 degrees. In the second case, when alpha was 240 degrees, angle ACB was also 120 degrees. Wait, that's strange. How?Wait, when we set alpha to 120 degrees, we placed point B at (-R/2, R√3/2 ), resulting in point C at (R/2, R√3/2 ). Angle ACB was 120 degrees. When we set alpha to 240 degrees, point B was at (-R/2, -R√3/2 ), resulting in point C at (R/2, -R√3/2 ). Angle ACB was also 120 degrees. So regardless of the angle between OA and OB being 120 or 240 degrees, angle ACB is 120 degrees. So why did we earlier think it could be 60 degrees? Because of the central angle. Wait, perhaps the key here is that in both configurations, angle ACB is 120 degrees. So maybe the answer is 120 degrees. But where did the earlier reasoning go wrong?The central angle AOB is 120 degrees. The inscribed angle ACB should be half the measure of the arc AB that it subtends. If point C is not on the arc AB, then angle ACB is half the central angle. But in our case, point C is on the arc AB, so angle ACB is half the measure of the opposite arc. Wait, let me think again. The measure of an inscribed angle is equal to half the measure of its intercepted arc. The intercepted arc is the arc that is inside the angle. In our case, angle ACB is formed by points A, C, B. The intercepted arc is the arc AB that does not contain point C. Wait, no. The intercepted arc is the arc that is cut off by the angle. So if you have angle ACB, the intercepted arc is arc AB that is opposite to C. So if point C is on the circumcircle, then the intercepted arc is the one that is not containing C. In our coordinate system, point C is on the minor arc AB (since OA and OB are 120 degrees apart, and OC is OA + OB). Therefore, the intercepted arc for angle ACB is the major arc AB, which is 240 degrees. Therefore, angle ACB is 240 / 2 = 120 degrees. Therefore, the correct answer is 120 degrees. But then why did the initial reasoning using the central angle say it should be 60 degrees? Because that reasoning assumed that angle ACB subtends the same arc as the central angle AOB. But in reality, since point C is located such that OA + OB = OC, point C is on the circumcircle but not diametrically opposed. Therefore, the intercepted arc is different. So the mistake was assuming that angle ACB subtends the same arc as the central angle AOB. But in reality, angle ACB subtends the opposite arc. Therefore, the measure of angle ACB is half of 240 degrees, which is 120 degrees. Therefore, the correct answer is 120 degrees. But this contradicts the previous coordinate calculation where when alpha was 120 degrees, the angle ACB was 120 degrees, but the central angle was also 120 degrees. Wait, no. The central angle AOB is 120 degrees. The intercepted arc AB is 240 degrees (since point C is on the minor arc AB), so angle ACB is 120 degrees. So yes, this makes sense. Therefore, the correct answer is 120 degrees. But wait, let's verify with another example. Suppose OA and OB are 60 degrees apart. Then OA + OB would have a magnitude greater than R, unless they are pointing in specific directions. Wait, but in our problem, OA + OB must have magnitude R. So only when the angle between OA and OB is 120 degrees or 240 degrees would |OA + OB| = R. Wait, when the angle between OA and OB is 120 degrees, |OA + OB| = R. Similarly, if the angle is 240 degrees, which is equivalent to -120 degrees, then |OA + OB| is also R. Therefore, in both cases, angle ACB is 120 degrees. Therefore, the answer should be 120 degrees. But now I'm confused because different methods gave different results. However, the coordinate calculation clearly shows angle ACB as 120 degrees. Therefore, the correct answer must be 120 degrees. Therefore, I must have made a mistake in my initial reasoning where I thought the inscribed angle is half the central angle. The key is to realize that point C is located such that it's on the circumcircle, but its position relative to arc AB affects which arc angle ACB subtends. Hence, the correct answer is 120 degrees, and the boxed answer is boxed{120^circ}.Wait, but let's check again. Maybe the problem is in Chinese or another language, and the original problem might have a figure. But since we don't have a figure, we have to rely on the vector equation. Given that OA + OB = OC, and O is the circumcenter. So, in vector terms, this means that point C is the vector sum of points A and B with respect to O. If we consider O as the origin, then the position vectors of A, B, and C satisfy A + B = C. Given that |A| = |B| = |C| = R. So, in such a case, the triangle formed by points A, B, and C has the property that C = A + B, with all three points on a circle of radius R centered at O. This is similar to a triangle inscribed in a circle where one vertex is the sum of the other two vertices as vectors. This configuration is known in geometry, and it's related to equilateral triangles or specific angles. For example, if A, B, and C form an equilateral triangle, then rotating a vector by 60 degrees and adding it would give the third vector. But in our case, the angle between OA and OB is 120 degrees, leading to OC being their sum. But regardless, according to the coordinate analysis, angle ACB is 120 degrees. Therefore, after careful consideration and verifying with coordinates, the answer should be 120 degrees. Final Answerboxed{120^circ}</think>Given that ( O ) is the circumcenter of (triangle ABC) and the vector equation (overrightarrow{OA} + overrightarrow{OB} = overrightarrow{OC}) holds, we need to determine the measure of (angle ACB).1. Understanding the Circumcenter: The circumcenter ( O ) is equidistant from all vertices of the triangle, so ( OA = OB = OC = R ), where ( R ) is the circumradius.2. Vector Analysis: The given vector equation implies that the vector (overrightarrow{OC}) is the resultant of (overrightarrow{OA}) and (overrightarrow{OB}). This can be interpreted geometrically using the properties of vectors and the parallelogram law of addition.3. Magnitude Calculation: - The magnitudes of the vectors (overrightarrow{OA}), (overrightarrow{OB}), and (overrightarrow{OC}) are equal to ( R ). - Using the dot product to find the angle between (overrightarrow{OA}) and (overrightarrow{OB}): [ |overrightarrow{OA} + overrightarrow{OB}|^2 = |overrightarrow{OC}|^2 ] [ |overrightarrow{OA}|^2 + |overrightarrow{OB}|^2 + 2 overrightarrow{OA} cdot overrightarrow{OB} = |overrightarrow{OC}|^2 ] [ R^2 + R^2 + 2 overrightarrow{OA} cdot overrightarrow{OB} = R^2 ] [ 2R^2 + 2R^2 cos theta = R^2 ] [ 2 + 2 cos theta = 1 ] [ cos theta = -frac{1}{2} ] [ theta = 120^circ ] - The angle between (overrightarrow{OA}) and (overrightarrow{OB}) is ( 120^circ ).4. Inscribed Angle Calculation: - The central angle (angle AOB) is ( 120^circ ). - The inscribed angle (angle ACB) subtends the arc ( AB ). Since the central angle is ( 120^circ ), the inscribed angle is half the measure of the arc that it subtends. - However, since the vector addition places point ( C ) such that (overrightarrow{OC} = overrightarrow{OA} + overrightarrow{OB}), point ( C ) is on the circumcircle and the inscribed angle (angle ACB) subtends the major arc ( AB ), which is ( 240^circ ).5. Final Calculation: - The inscribed angle (angle ACB) is half of the major arc ( AB ): [ angle ACB = frac{240^circ}{2} = 120^circ ]Thus, the measure of (angle ACB) is (boxed{120^circ}).,<think>Okay, let's try to solve this problem. So, we have triangle ABC with circumcenter O, and the vector equation OA + OB = OC. We need to find the measure of angle ACB. Hmm, vectors and circumcenters... Let me recall what I know.First, the circumcenter O is the intersection of the perpendicular bisectors of the triangle's sides. It's equidistant from all three vertices, so OA = OB = OC in length. That's important because vectors have both magnitude and direction, but here the magnitudes are equal. The equation given is vector OA plus vector OB equals vector OC. So, maybe I can use vector addition properties here.Let me denote vectors OA, OB, and OC as vectors from the origin, but wait, actually in this problem, O is the circumcenter, so all vectors are originating from point O. So OA is the vector from O to A, OB from O to B, and OC from O to C. So the equation is OA + OB = OC. That means if I add the vectors OA and OB, the resultant vector is OC. Since OA, OB, and OC are all radii of the circumcircle, their magnitudes are equal. Let's denote |OA| = |OB| = |OC| = R, where R is the circumradius.So, OA + OB = OC. Let's square both sides to use the property of vectors. The square of a vector's magnitude is the dot product of the vector with itself. So,(OA + OB) · (OA + OB) = OC · OCExpanding the left side:OA · OA + 2 OA · OB + OB · OB = OC · OCSince |OA| = |OB| = |OC| = R, we have:R² + 2 OA · OB + R² = R²Simplify:2 R² + 2 OA · OB = R²Subtract R² from both sides:R² + 2 OA · OB = 0So,2 OA · OB = -R²Divide both sides by 2:OA · OB = -R² / 2Now, the dot product OA · OB is also equal to |OA||OB|cosθ, where θ is the angle between vectors OA and OB. Since |OA| = |OB| = R, we have:R * R * cosθ = -R² / 2Simplify:R² cosθ = -R² / 2Divide both sides by R² (assuming R ≠ 0, which it isn't because it's a triangle):cosθ = -1/2So θ, the angle between OA and OB, is arccos(-1/2). The angle whose cosine is -1/2 is 120 degrees. So, angle AOB is 120 degrees.But angle AOB is the central angle corresponding to the arc AB. In a circle, the central angle is twice the inscribed angle subtended by the same arc. So, the inscribed angle ACB would be half of 120 degrees, which is 60 degrees.Wait, let me confirm that. The central angle is 120 degrees, so the inscribed angle would be half of that, which is 60 degrees. So angle ACB is 60 degrees. Is that right?But let me check if angle ACB is indeed subtended by arc AB. In triangle ABC, angle at C (angle ACB) is subtended by arc AB. Yes, because in a circle, the inscribed angle at point C subtended by arc AB is half the measure of the central angle subtended by arc AB. Since the central angle is 120 degrees, the inscribed angle would be 60 degrees. Therefore, angle ACB is 60 degrees.But wait, let me make sure there's no mistake here. Maybe I need to verify using another approach.Alternatively, since OA + OB = OC, we can use vector analysis. Let me consider coordinates. Let's place the circumcenter O at the origin for simplicity. Then vectors OA, OB, OC are position vectors of points A, B, C. So OA is vector A, OB is vector B, OC is vector C. Then the equation becomes A + B = C.Given that |A| = |B| = |C| = R.So, C = A + B. Then |C| = |A + B|. But |C| = R, and |A + B| = sqrt(|A|² + |B|² + 2A·B) = sqrt(R² + R² + 2A·B). But since |A + B| = |C| = R, we have:sqrt(2R² + 2A·B) = RSquaring both sides:2R² + 2A·B = R²Which simplifies to A·B = -R²/2, which is the same result as before. Then the angle between A and B (which is angle AOB) is 120 degrees. So same conclusion.Then angle ACB is half of angle AOB, so 60 degrees. That seems consistent.Alternatively, maybe we can use coordinate geometry. Let me try placing point O at the origin. Let’s assign coordinates such that vector OA is along the x-axis. Let’s say point A is at (R, 0). Then vector OA is (R, 0). Let vector OB be (R cosθ, R sinθ), since |OB| = R. Then vector OC = OA + OB = (R + R cosθ, R sinθ). But since |OC| = R, we can write:(R + R cosθ)² + (R sinθ)² = R²Expanding:R²(1 + 2 cosθ + cos²θ) + R² sin²θ = R²Factor out R²:R²[1 + 2 cosθ + cos²θ + sin²θ] = R²Since cos²θ + sin²θ = 1:R²[1 + 2 cosθ + 1] = R²Simplify:R²[2 + 2 cosθ] = R²Divide both sides by R²:2 + 2 cosθ = 1Subtract 1:1 + 2 cosθ = 0So 2 cosθ = -1 => cosθ = -1/2 => θ = 120 degrees. So angle AOB is 120 degrees. Then angle ACB is 60 degrees. Yep, same result.Therefore, angle ACB is 60 degrees, which is π/3 radians. So the answer should be 60°, which in boxed form is boxed{60^circ}.But let me check once more. Suppose angle AOB is 120 degrees, then the arc AB is 120 degrees. The inscribed angle over the same arc AB would be 60 degrees. So angle ACB is 60 degrees, yes. Alternatively, in triangle ABC, if the central angle is 120 degrees, then the triangle formed by the points A, B, C would have certain properties. For instance, in such a case, triangle ABC would have angle at C equal to 60 degrees. That seems correct.Alternatively, maybe using the law of cosines on triangle ABC. Wait, but we might not know the sides. Wait, but perhaps triangle OAC and OBC?Wait, OA, OB, OC are all radii. If OA + OB = OC, then in coordinate terms, with O at the origin, C has coordinates (A_x + B_x, A_y + B_y). Then AC is the vector from A to C, which is (A_x + B_x - A_x, A_y + B_y - A_y) = (B_x, B_y). Similarly, BC is (A_x + B_x - B_x, A_y + B_y - B_y) = (A_x, A_y). So vectors AC and BC are just vectors OB and OA, respectively. Wait, that might be helpful.So, vectors AC = OC - OA = (OA + OB) - OA = OB. Similarly, vectors BC = OC - OB = (OA + OB) - OB = OA. So AC = OB and BC = OA. Since OA and OB have the same magnitude, then AC and BC have the same magnitude as OA and OB. Therefore, AC = BC, so triangle ABC is isoceles with AC = BC. Therefore, angle at C is angle ACB, which is the vertex angle of an isoceles triangle with sides AC = BC.But wait, if AC = BC, then angles at A and B are equal. Wait, but we found angle ACB is 60 degrees. If it's isoceles with two equal sides AC and BC, then the base angles at A and B are equal. But if angle at C is 60 degrees, then the triangle is equilateral. Wait, but that would require all sides equal. But OA, OB, OC are radii, but AC and BC being equal to OA and OB would mean that AC = BC = R, but OA = OB = R as well. Wait, so if OA = AC = BC = R, then points A and B are at distance R from O and from C. Hmm, maybe triangle AOC and BOC?Wait, OA = OC? No, because OC is OA + OB. If OA and OB are vectors of length R at 120 degrees, then OC is the resultant vector. The magnitude of OC is R, as given. So |OA + OB| = R. So OA and OB are vectors with magnitude R, angle 120 degrees between them, and their sum has magnitude R.But in such a case, if OA and OB are at 120 degrees, then |OA + OB| can be calculated. Wait, |OA + OB|² = |OA|² + |OB|² + 2 OA · OB = R² + R² + 2*(R*R*cos 120°) = 2 R² + 2 R²*(-1/2) = 2 R² - R² = R². So |OA + OB| = R, which matches |OC| = R. So that checks out.So, in this configuration, points A, B, C are on the circumcircle centered at O. OA and OB are separated by 120 degrees, and OC is their vector sum. So the triangle ABC is such that C is located at the position OA + OB. So in coordinates, if OA is (R,0) and OB is (R cos 120°, R sin 120°), which is (-R/2, R√3/2). Then OC = OA + OB = (R - R/2, 0 + R√3/2) = (R/2, R√3/2). So point C is at (R/2, R√3/2). Now, let's find angle ACB.Point A is at (R, 0), point B is at (-R/2, R√3/2), and point C is at (R/2, R√3/2). Let's compute vectors CA and CB.Vector CA = A - C = (R - R/2, 0 - R√3/2) = (R/2, -R√3/2)Vector CB = B - C = (-R/2 - R/2, R√3/2 - R√3/2) = (-R, 0)Now, angle ACB is the angle between vectors CA and CB. Let's compute the angle between (R/2, -R√3/2) and (-R, 0).The formula for the angle θ between vectors u and v is:cosθ = (u · v) / (|u||v|)Compute the dot product:u · v = (R/2)(-R) + (-R√3/2)(0) = -R²/2 + 0 = -R²/2|u| = sqrt((R/2)^2 + (-R√3/2)^2) = sqrt(R²/4 + 3R²/4) = sqrt(R²) = R|v| = sqrt((-R)^2 + 0^2) = RThus,cosθ = (-R²/2) / (R * R) = -1/2Therefore, θ = arccos(-1/2) = 120 degrees. Wait, but that's the angle between vectors CA and CB, which is angle ACB. Wait, but earlier we thought angle ACB is 60 degrees. There's a contradiction here.Hold on, this is confusing. According to the coordinate geometry approach, angle ACB is 120 degrees, but according to the central angle and inscribed angle relationship, it's 60 degrees. Which one is correct?Wait, maybe I made a mistake in interpreting the angle. Let me check the coordinates again.Point A: (R, 0)Point B: (-R/2, R√3/2)Point C: (R/2, R√3/2)So, plotting these points:- A is at (R, 0), which is on the positive x-axis.- B is at (-R/2, R√3/2), which is 120 degrees from the x-axis.- C is at (R/2, R√3/2), which is 60 degrees from the x-axis.So, triangle ABC has points:A: (R, 0)B: (-R/2, R√3/2)C: (R/2, R√3/2)Let me sketch this mentally. Point C is in the upper half of the circle, between A and B. Wait, point C is actually at (R/2, R√3/2), which is 60 degrees above the x-axis. Point B is at (-R/2, R√3/2), which is 120 degrees above the negative x-axis. So, points A, C are on the right side, and B is on the left side.So, triangle ABC is formed by these three points. Let me compute the angle at C. So angle ACB is the angle at point C between points A and B.Wait, in coordinates, point C is (R/2, R√3/2). Let me compute the vectors from C to A and from C to B.Vector CA: A - C = (R - R/2, 0 - R√3/2) = (R/2, -R√3/2)Vector CB: B - C = (-R/2 - R/2, R√3/2 - R√3/2) = (-R, 0)So vectors CA and CB are (R/2, -R√3/2) and (-R, 0). The angle between these vectors is computed as:cosθ = ( (R/2)(-R) + (-R√3/2)(0) ) / ( |CA| |CB| )Which is (-R²/2) / ( R * R ) = -1/2, so θ = 120 degrees. So angle ACB is 120 degrees? But this contradicts the previous conclusion of 60 degrees. So which is correct?Wait, but according to the central angle, angle AOB is 120 degrees, and the inscribed angle over the same arc AB should be half of that, so 60 degrees. But according to coordinates, angle at C is 120 degrees. So there's a discrepancy here. Where is the mistake?Wait, perhaps the arc AB is actually the other arc? In a circle, the central angle can be the minor arc or the major arc. The inscribed angle depends on which arc we take. The central angle here is 120 degrees, which is the minor arc AB. Then the inscribed angle would be 60 degrees. But in our coordinate setup, angle ACB is 120 degrees. How is this possible?Wait, perhaps the points are arranged such that angle ACB is actually the inscribed angle over the major arc AB. Because in the circle, if the central angle is 120 degrees, the major arc AB is 240 degrees. Then the inscribed angle over the major arc would be half of 240, which is 120 degrees. So angle ACB could be 120 degrees if it's subtended by the major arc AB.But in order for angle ACB to be subtended by the major arc AB, point C must lie on the opposite side of the circle from the major arc AB. Wait, in our coordinate system, point C is at (R/2, R√3/2), which is 60 degrees above the x-axis. Points A and B are at (R, 0) and (-R/2, R√3/2). So arc AB going through the upper half of the circle would be the minor arc of 120 degrees, and the major arc AB going through the lower half would be 240 degrees. But point C is in the upper half, so angle ACB would be subtended by the minor arc AB, leading to 60 degrees. But according to coordinates, angle ACB is 120 degrees. Contradiction.Wait, this is confusing. Let me check the coordinates again. Maybe I made a mistake in calculation.Calculating angle at point C (coordinates (R/2, R√3/2)) between points A (R,0) and B (-R/2, R√3/2).First, compute the lengths of sides AC, BC, and AB.AC: distance between A and C.A is (R, 0), C is (R/2, R√3/2)Distance AC: sqrt( (R - R/2)^2 + (0 - R√3/2)^2 ) = sqrt( (R/2)^2 + ( - R√3/2 )^2 ) = sqrt( R²/4 + 3R²/4 ) = sqrt(R²) = RSimilarly, BC: distance between B and C.B is (-R/2, R√3/2), C is (R/2, R√3/2)Distance BC: sqrt( ( - R/2 - R/2 )^2 + ( R√3/2 - R√3/2 )^2 ) = sqrt( (-R)^2 + 0 ) = sqrt(R²) = RAB: distance between A and B.A is (R, 0), B is (-R/2, R√3/2)Distance AB: sqrt( (R + R/2)^2 + (0 - R√3/2)^2 ) = sqrt( (3R/2)^2 + ( - R√3/2 )^2 ) = sqrt( 9R²/4 + 3R²/4 ) = sqrt(12R²/4) = sqrt(3R²) = R√3So triangle ABC has sides AC = BC = R, AB = R√3. So it's an isoceles triangle with two sides equal to R and base R√3. Let's compute angle at C.Using the law of cosines on triangle ABC:AB² = AC² + BC² - 2 AC * BC * cos(angle ACB)( R√3 )² = R² + R² - 2 * R * R * cos(angle ACB )3 R² = 2 R² - 2 R² cos(angle ACB )Subtract 2 R²:R² = -2 R² cos(angle ACB )Divide by -2 R²:-1/2 = cos(angle ACB )Thus, angle ACB = arccos(-1/2) = 120 degrees.But according to the central angle and inscribed angle relationship, angle ACB should be half of angle AOB. But angle AOB is 120 degrees, so inscribed angle should be 60 degrees. Why the discrepancy?Wait, perhaps the inscribed angle theorem states that the angle is half the central angle subtended by the same arc. But in this case, angle ACB is subtended by arc AB. However, in our configuration, point C is not on the circumcircle. Wait, no, all points A, B, C are on the circumcircle, because OA, OB, OC are radii. Wait, but in this case, points A, B, C are on the circle centered at O with radius R.But according to the coordinates, angle ACB is 120 degrees, which is the same as the central angle of 120 degrees. Wait, but that's impossible, because inscribed angle should be half. Wait, unless point C is diametrically opposite to the midpoint of arc AB?Wait, in our coordinate system, points A, B, C are all on the circle. Let me check their positions again. Point A is (R, 0), point C is (R/2, R√3/2), which is 60 degrees above the x-axis. Point B is (-R/2, R√3/2), which is 120 degrees above the negative x-axis.So, arc from A to B passing through the upper half of the circle is 120 degrees (central angle). Then the inscribed angle over that arc should be 60 degrees. But in reality, angle ACB is 120 degrees. So this seems contradictory.Wait, maybe the problem is that angle ACB is not subtended by arc AB but by another arc?Wait, angle ACB is formed by points A, C, B. In a circle, the inscribed angle at point C subtended by arc AB. But depending on the position of point C, the angle can be either the minor arc or the major arc. Since point C is located such that OC = OA + OB, which in coordinates puts it in the upper half between A and B, but according to the computation, angle ACB is 120 degrees. Wait, but according to the law of cosines, angle ACB is 120 degrees, which is correct based on the coordinates. So the previous reasoning that angle ACB is half of angle AOB must be wrong here. So why is that?Wait, perhaps because point C is not located on the arc AB that we are considering. Let me think. The central angle AOB is 120 degrees. If point C is located such that OC is the vector sum of OA and OB, then in our coordinate system, point C is located at (R/2, R√3/2), which is in the same circle. However, the arc from A to B passing through C is actually 240 degrees, since from A (0 degrees) to C (60 degrees) to B (120 degrees from negative x-axis, which is 180 - 120 = 60 degrees above negative x-axis, so 180 - 60 = 120 degrees from positive x-axis). Wait, maybe this is getting too confusing.Alternatively, perhaps the key lies in the fact that in this specific configuration, due to the vector addition, angle ACB is actually 120 degrees. Therefore, the earlier reasoning about inscribed angles might not apply here because point C is not located on the minor arc AB but on the major arc AB. Wait, inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. If angle ACB is subtended by arc AB, then the measure of angle ACB is half the measure of arc AB. But in our case, angle ACB is 120 degrees, so the arc AB must measure 240 degrees. But the central angle AOB is 120 degrees. So which is it?Wait, the central angle AOB is 120 degrees, which corresponds to the minor arc AB. The major arc AB would then be 360 - 120 = 240 degrees. So if angle ACB is subtended by the major arc AB, then the inscribed angle would be 240 / 2 = 120 degrees, which matches our calculation. Therefore, angle ACB is subtended by the major arc AB, meaning that point C is located on the opposite side of the circle from the minor arc AB.But in our coordinate system, point C is at (R/2, R√3/2), which is in the upper half, same as point B. However, the major arc AB would go around the lower half of the circle. So perhaps in this configuration, point C is indeed located such that angle ACB is subtended by the major arc AB. Let me confirm.If we traverse the circle from A to B through the lower half (the longer path), that's the major arc AB of 240 degrees. The inscribed angle subtended by this arc would indeed be 120 degrees. However, for the inscribed angle to subtend the major arc, point C must lie on the opposite side of the circle from the major arc. Wait, no. The inscribed angle subtended by an arc depends on where the point is located. If the point is on one side, it subtends the arc on the opposite side.Therefore, if point C is located on the major arc AB, then angle ACB would subtend the minor arc AB. But if point C is located on the minor arc AB, angle ACB would subtend the major arc AB. Wait, this is getting confusing. Let me recall the exact statement.In a circle, the measure of an inscribed angle is half the measure of its intercepted arc. The intercepted arc is the arc that is inside the angle and between the two sides of the angle. So, if we have points A, B, C on the circle, angle ACB intercepts the arc AB that does not contain point C. Therefore, if point C is on the major arc AB, then angle ACB intercepts the minor arc AB, and vice versa.Wait, yes. So if point C is on the major arc AB (the longer arc), then angle ACB intercepts the minor arc AB (the shorter arc). Conversely, if point C is on the minor arc AB, angle ACB intercepts the major arc AB.In our case, points A, B, C are such that OA + OB = OC. In the coordinate system, point C is in the upper half between A and B. Wait, but with our coordinates, point C is at (R/2, R√3/2), which is 60 degrees from the x-axis. Point A is at 0 degrees, point B is at 120 degrees from the x-axis? Wait, no. Point B is at (-R/2, R√3/2), which is actually 150 degrees from the positive x-axis? Wait, let's compute the angle of point B.Coordinates of B: (-R/2, R√3/2). The angle θ from positive x-axis is arctan( (R√3/2) / (-R/2) ) = arctan(-√3). Since the x-coordinate is negative and y-coordinate is positive, it's in the second quadrant. arctan(-√3) is -60 degrees, but since it's in the second quadrant, the angle is 180 - 60 = 120 degrees. So point B is at 120 degrees from the positive x-axis.Point C is at (R/2, R√3/2), which is in the first quadrant. The angle here is arctan( (R√3/2) / (R/2) ) = arctan(√3) = 60 degrees. So point C is at 60 degrees, point A is at 0 degrees, point B is at 120 degrees. So the minor arc AB is from 0 degrees to 120 degrees, which is 120 degrees. The major arc AB is from 120 degrees back to 0 degrees the long way around, which is 240 degrees.Now, angle ACB is at point C (60 degrees). To find the intercepted arc by angle ACB, we need to see which arc is opposite to point C. The intercepted arc AB is the arc that does not contain point C. Since point C is at 60 degrees, between A (0 degrees) and B (120 degrees), the arc AB that does not contain C is the major arc AB going from A (0 degrees) through 180 degrees to B (120 degrees). Wait, actually, if point C is between A and B on the minor arc AB, then the intercepted arc AB for angle ACB would be the major arc AB. Because the intercepted arc is the one not containing the vertex of the angle.Therefore, angle ACB intercepts the major arc AB (240 degrees) and thus the measure of angle ACB is half of 240 degrees, which is 120 degrees. Therefore, angle ACB is 120 degrees. But wait, this contradicts the initial conclusion based on the central angle. Wait, no. The central angle AOB is 120 degrees, which is the minor arc AB. The major arc AB is 240 degrees. So angle ACB, which intercepts the major arc AB, is 120 degrees. Therefore, the correct measure is 120 degrees.But earlier, when using the law of cosines on triangle ABC, we found angle ACB to be 120 degrees, which aligns with this. However, initially, I thought it was 60 degrees, which was a mistake because I incorrectly assumed that the inscribed angle subtends the same arc as the central angle, but in reality, depending on the position of point C, it can subtend the opposite arc.Therefore, the correct answer is 120 degrees. But the problem says "angle ACB". So which is correct?Wait, let's re-examine everything. The problem states that O is the circumcenter and OA + OB = OC. We derived that angle AOB is 120 degrees. Then, depending on where point C is, angle ACB is either 60 degrees or 120 degrees. However, based on the coordinate calculation and law of cosines, it's 120 degrees. So why is there a discrepancy?Ah, I think the mistake was in the initial assumption that angle ACB is subtended by the same arc AB as angle AOB. But actually, angle ACB is subtended by the major arc AB, not the minor arc. Since point C is not on the minor arc AB (it's on the major arc), the inscribed angle is half the measure of the major arc AB, which is 240 degrees. Therefore, angle ACB is 120 degrees.Therefore, the correct answer is 120 degrees, which conflicts with the initial thought. But according to the coordinate geometry and law of cosines, it's 120 degrees. So where did the earlier reasoning go wrong?Wait, another way: since OA + OB = OC, then point C is the fourth vertex of the rhombus formed by OA and OB. But since OA and OB have the same length and the angle between them is 120 degrees, the figure formed by OA, OB, and OC is not a rhombus because OC is equal in length to OA and OB. Wait, OA + OB = OC, and |OA| = |OB| = |OC|. In such a case, the triangle formed by vectors OA, OB, and OC is an equilateral triangle? Wait, no. If OA and OB are vectors with 120 degrees between them, and their sum OC is also of the same length, then the triangle formed by OA, OB, and OC is such that OA + OB = OC. But this is only possible if the angle between OA and OB is 120 degrees, as we found.But this seems to form a triangle where OC is the resultant vector. However, in the coordinate system, the angle ACB came out to be 120 degrees. Therefore, the answer should be 120 degrees. So the initial mistake was assuming that angle ACB is subtended by the minor arc AB, but in reality, it's subtended by the major arc AB, leading to 120 degrees.Therefore, the correct answer is 120 degrees. But wait, this conflicts with the problem's context. Let me check once more.Wait, in the problem statement, they say "circumcenter O of triangle ABC", so triangle ABC must be non-degenerate. If angle ACB is 120 degrees, that's acceptable. But we need to confirm.Given the vector equation OA + OB = OC, which in coordinates leads to angle ACB of 120 degrees. But when we used the central angle and inscribed angle theorem, we have to be careful which arc is intercepted. Since point C is located such that OC = OA + OB, which in coordinates is in the first quadrant, between OA and OB (which is actually not between OA and OB in terms of central angles). Wait, point C is not between OA and OB in terms of position.Wait, OA is at 0 degrees, OB is at 120 degrees, and OC is at 60 degrees. So point C is at 60 degrees, which is between OA (0 degrees) and OB (120 degrees). However, the intercepted arc for angle ACB would be the arc AB that doesn't contain point C. Since point C is at 60 degrees, between A (0 degrees) and B (120 degrees), the arc AB that doesn't contain C is the major arc AB passing through 180 degrees. Hence, angle ACB is half of 240 degrees, which is 120 degrees. Therefore, angle ACB is 120 degrees.Therefore, the correct answer is 120 degrees, which should be boxed as boxed{120^circ}. But I need to verify this once more.Wait, going back to the vector approach. If OA, OB, and OC have the same magnitude, and OA + OB = OC, then the triangle formed by OA, OB, and OC is such that the sum of OA and OB is OC. In vector terms, this is only possible if the angle between OA and OB is 120 degrees. Then, in triangle ABC, using the law of cosines gave angle ACB as 120 degrees. Therefore, the answer should be 120 degrees.But wait, the answer key for similar problems often has 60 degrees. Wait, perhaps I made a mistake in interpreting the angle. Let me check once more.Wait, in the coordinate system, angle ACB is at point C (60 degrees, R√3/2). Point A is (R, 0), point B is (-R/2, R√3/2). Let me compute the angles using the tangent method.The angle at point C can be calculated by finding the slopes of lines CA and CB.Slope of CA: (0 - R√3/2) / (R - R/2) = (-R√3/2) / (R/2) = -√3Slope of CB: (R√3/2 - R√3/2) / (-R/2 - R/2) = 0 / (-R) = 0So line CB is horizontal (slope 0), line CA has slope -√3. The angle between a horizontal line and a line with slope -√3. The angle whose tangent is |slope| is √3, which corresponds to 60 degrees. However, since the slope is negative, the angle is measured below the horizontal line. Therefore, the angle between CA and CB is 180 - 60 = 120 degrees. So yes, angle ACB is 120 degrees.Therefore, after multiple methods, the angle ACB is 120 degrees. Therefore, the initial mistake was in assuming that the inscribed angle is subtended by the minor arc, but in reality, it's subtended by the major arc, leading to 120 degrees. Hence, the answer is 120 degrees.Final Answerboxed{120^circ}</think>Given triangle ( triangle ABC ) with circumcenter ( O ), we know that ( overrightarrow{OA} + overrightarrow{OB} = overrightarrow{OC} ). We need to find the measure of ( angle ACB ).1. Vector Analysis: - Since ( O ) is the circumcenter, ( |OA| = |OB| = |OC| = R ). - The given vector equation ( overrightarrow{OA} + overrightarrow{OB} = overrightarrow{OC} ) implies: [ (overrightarrow{OA} + overrightarrow{OB}) cdot (overrightarrow{OA} + overrightarrow{OB}) = overrightarrow{OC} cdot overrightarrow{OC} ] - Expanding and simplifying using the dot product: [ R^2 + 2 overrightarrow{OA} cdot overrightarrow{OB} + R^2 = R^2 ] [ 2R^2 + 2 overrightarrow{OA} cdot overrightarrow{OB} = R^2 ] [ overrightarrow{OA} cdot overrightarrow{OB} = -frac{R^2}{2} ] - The dot product ( overrightarrow{OA} cdot overrightarrow{OB} = R^2 cos theta ), where ( theta ) is the angle between ( overrightarrow{OA} ) and ( overrightarrow{OB} ). Thus: [ cos theta = -frac{1}{2} implies theta = 120^circ ] - The central angle ( angle AOB = 120^circ ).2. Inscribed Angle Theorem: - The inscribed angle ( angle ACB ) subtended by the major arc ( AB ) (which is ( 240^circ )) is half the measure of the central angle subtended by the same arc: [ angle ACB = frac{240^circ}{2} = 120^circ ]3. Coordinate Geometry Verification: - Placing ( O ) at the origin, coordinates of ( A ), ( B ), and ( C ) confirm the vectors and angle calculations. - Using the law of cosines in ( triangle ABC ), the angle ( angle ACB ) is found to be ( 120^circ ).Thus, the measure of ( angle ACB ) is ( boxed{120^circ} ).]

🔑:6. frac{2 pi}{3}.From overrightarrow{O A}+overrightarrow{O B}=overrightarrow{O C}, we know that quadrilateral O A C B is a parallelogram.Also, |O A|=|O B|=|O C|, so, square O A C B is a rhombus, and angle O A C=angle O B C=frac{pi}{3}.Thus, angle A C B=frac{2 pi}{3}.

🧠:[<think>Okay, so I have this problem about a parabola and a straight line. Let me try to break it down step by step. First, part (1) says: If the slope k=2, calculate the length of the chord that the line cuts from the parabola. The parabola is given by y² = 12x, and the line has a slope of 2 passing through the point (0, -1). Alright, so I need to find the equation of the line first. Since it has a slope of 2 and passes through (0, -1), the equation should be in the form y = mx + b. Here, m is 2, and since when x=0, y=-1, the y-intercept b is -1. So the equation is y = 2x - 1.Next, I need to find where this line intersects the parabola y² = 12x. To do that, I can substitute y from the line equation into the parabola equation. Let's do that:Substitute y = 2x - 1 into y² = 12x:(2x - 1)² = 12xExpand the left side:4x² - 4x + 1 = 12xBring all terms to one side:4x² - 4x + 1 - 12x = 0Combine like terms:4x² - 16x + 1 = 0So, quadratic equation in x: 4x² - 16x + 1 = 0.To find the solutions for x, we can use the quadratic formula. The quadratic is ax² + bx + c = 0, so here a=4, b=-16, c=1.Discriminant D = b² - 4ac = (-16)² - 4*4*1 = 256 - 16 = 240.So, sqrt(D) = sqrt(240) = sqrt(16*15) = 4*sqrt(15).Thus, the solutions for x are:x = [16 ± 4√15]/(2*4) = [16 ± 4√15]/8 = [4 ± √15]/2.So, x coordinates of the intersection points are (4 + √15)/2 and (4 - √15)/2.Now, to find the corresponding y coordinates, plug back into y = 2x - 1.First point: x = (4 + √15)/2y = 2*(4 + √15)/2 - 1 = (4 + √15) - 1 = 3 + √15.Second point: x = (4 - √15)/2y = 2*(4 - √15)/2 - 1 = (4 - √15) - 1 = 3 - √15.So, the two intersection points are:Point A: ((4 + √15)/2, 3 + √15)Point B: ((4 - √15)/2, 3 - √15)Now, to find the length of the chord AB, we can use the distance formula between these two points.Let me denote the coordinates as:A: (x₁, y₁) = ((4 + √15)/2, 3 + √15)B: (x₂, y₂) = ((4 - √15)/2, 3 - √15)Distance AB = sqrt[(x₂ - x₁)² + (y₂ - y₁)²]Calculate x₂ - x₁:[(4 - √15)/2] - [(4 + √15)/2] = [4 - √15 - 4 - √15]/2 = (-2√15)/2 = -√15Similarly, y₂ - y₁:(3 - √15) - (3 + √15) = -2√15So, Distance AB = sqrt[(-√15)² + (-2√15)²] = sqrt[15 + 4*15] = sqrt[15 + 60] = sqrt[75] = sqrt[25*3] = 5*sqrt(3)So, the length of the chord is 5√3.Wait, let me check that again. Because sqrt(75) is indeed 5√3. Yes, 75=25*3. So that part is correct. Let me verify the coordinates again.Original equation: y = 2x - 1.For x = (4 + √15)/2, y = 2*(4 + √15)/2 -1 = 4 + √15 -1 = 3 + √15. Correct.Similarly, for the other x, y = 3 - √15. Correct.Then differences in coordinates: x difference is -√15, y difference is -2√15. Squaring those gives 15 and 60, sum 75. Square root of 75 is 5√3. So, yes, that seems right. Okay, so part (1) answer is 5√3.Now, part (2): If the line has only one intersection point with the parabola, determine the possible value(s) of the slope k.So, the line has slope k, passes through (0, -1), so equation is y = kx - 1. We need this line to intersect the parabola y² = 12x exactly once. That is, the system of equations has exactly one solution. So, substitute y = kx -1 into y² = 12x:(kx -1)^2 = 12xExpand left side:k²x² - 2kx +1 = 12xBring all terms to left:k²x² - (2k +12)x +1 = 0So, quadratic in x: k²x² - (2k +12)x +1 = 0For this quadratic to have exactly one solution, the discriminant must be zero.Discriminant D = [-(2k +12)]² - 4*k²*1 = (2k +12)² -4k²Compute this:First, expand (2k +12)^2: 4k² + 48k +144Then subtract 4k²: 4k² +48k +144 -4k² = 48k +144Set discriminant to zero:48k +144 = 0Solve for k:48k = -144k = -144/48 = -3So, the possible value of k is -3.Wait, let me verify this. Substitute back:If k = -3, then the line is y = -3x -1.Substitute into y² = 12x:(-3x -1)^2 = 9x² +6x +1 = 12xSo, 9x² +6x +1 -12x = 9x² -6x +1 =0Discriminant: (-6)^2 -4*9*1 = 36 -36 =0. So, yes, repeated root. Thus, the line is tangent to the parabola. So, k=-3 is the only value. So, part (2) answer is k=-3.Wait, but let me think again. The equation when substituting y=kx-1 into parabola gives discriminant 48k +144=0, so k=-3. That seems correct. So, yes, only k=-3. So, possible value is -3.So, summarizing:1) Length is 5√32) k=-3I think that's it. Let me just check if in part (1) my steps are correct. The substitution seems okay, solving quadratic, getting the x's, then y's, then distance. The distance squared is (√15)^2 + (2√15)^2 =15 +60=75, sqrt(75)=5√3. Correct.Another way to compute the length: since the line intersects the parabola, maybe using parametric equations or calculus, but since the first method worked, it's okay.For part (2), discriminant approach is standard for tangency. So, yes, k=-3. Okay, confident with these answers.Final Answer(1) The length of the chord is boxed{5sqrt{3}}.(2) The possible value of the slope ( k ) is boxed{-3}.</think>Given the parabola ( y^2 = 12x ) and a straight line with slope ( k ) passing through the point ( (0, -1) ):1. When the slope ( k = 2 ): - The equation of the line is ( y = 2x - 1 ). - Substitute ( y = 2x - 1 ) into ( y^2 = 12x ): [ (2x - 1)^2 = 12x ] - Expand and simplify: [ 4x^2 - 4x + 1 = 12x implies 4x^2 - 16x + 1 = 0 ] - Solve the quadratic equation using the quadratic formula: [ x = frac{16 pm sqrt{240}}{8} = frac{4 pm sqrt{15}}{2} ] - Find the corresponding ( y )-coordinates: [ y = 2 left( frac{4 + sqrt{15}}{2} right) - 1 = 3 + sqrt{15} ] [ y = 2 left( frac{4 - sqrt{15}}{2} right) - 1 = 3 - sqrt{15} ] - Calculate the distance between the points ( left( frac{4 + sqrt{15}}{2}, 3 + sqrt{15} right) ) and ( left( frac{4 - sqrt{15}}{2}, 3 - sqrt{15} right) ): [ text{Distance} = sqrt{ left( -sqrt{15} right)^2 + left( -2sqrt{15} right)^2 } = sqrt{15 + 60} = sqrt{75} = 5sqrt{3} ]2. When the line has only one intersection point with the parabola: - The equation of the line is ( y = kx - 1 ). - Substitute ( y = kx - 1 ) into ( y^2 = 12x ): [ (kx - 1)^2 = 12x ] - Expand and simplify: [ k^2x^2 - 2kx + 1 = 12x implies k^2x^2 - (2k + 12)x + 1 = 0 ] - Set the discriminant to zero for a single solution: [ (2k + 12)^2 - 4k^2 = 0 ] - Solve for ( k ): [ 4k^2 + 48k + 144 - 4k^2 = 0 implies 48k + 144 = 0 implies k = -3 ] Final Answer(1) The length of the chord is (boxed{5sqrt{3}}).(2) The possible value of the slope ( k ) is (boxed{-3}).,<think>Okay, so I need to solve two parts here. Let me start with part (1). The problem says: Given the parabola y² = 12x, and a straight line with slope k=2 passing through (0, -1), calculate the length of the chord that the line cuts from the parabola.First, I need to find the equation of the line. Since the slope is 2 and it passes through (0, -1), the equation should be y = 2x - 1. Wait, because the general form is y - y1 = k(x - x1). Here, (x1, y1) is (0, -1), so y - (-1) = 2(x - 0), which simplifies to y = 2x - 1. That seems right.Next, I need to find where this line intersects the parabola y² = 12x. So I can substitute y from the line equation into the parabola equation. Let's do that:(2x - 1)² = 12xExpanding the left side: (2x)^2 - 2*(2x)*(1) + 1² = 4x² - 4x + 1So the equation becomes:4x² - 4x + 1 = 12xLet's bring all terms to one side:4x² - 4x + 1 - 12x = 0Combine like terms:4x² - 16x + 1 = 0So, quadratic equation in x: 4x² -16x +1 =0I need to solve this quadratic for x. Let's use the quadratic formula:x = [16 ± sqrt(256 - 16)] / 8Wait, discriminant D = (-16)^2 - 4*4*1 = 256 -16 = 240So sqrt(240) = sqrt(16*15) = 4*sqrt(15)Therefore, x = [16 ± 4√15]/8 = [16/8] ± [4√15]/8 = 2 ± (√15)/2So the x-coordinates of the intersection points are 2 + (√15)/2 and 2 - (√15)/2.Now, let's find the corresponding y-coordinates using the line equation y = 2x -1.For x = 2 + (√15)/2:y = 2*(2 + √15/2) -1 = 4 + √15 -1 = 3 + √15For x = 2 - (√15)/2:y = 2*(2 - √15/2) -1 = 4 - √15 -1 = 3 - √15So the two points of intersection are:(2 + (√15)/2, 3 + √15) and (2 - (√15)/2, 3 - √15)Now, to find the length of the chord between these two points, we can use the distance formula.Let me denote the points as (x1, y1) and (x2, y2):x1 = 2 + √15/2, y1 = 3 + √15x2 = 2 - √15/2, y2 = 3 - √15First, compute the difference in x and y:Δx = x1 - x2 = [2 + √15/2] - [2 - √15/2] = √15Δy = y1 - y2 = [3 + √15] - [3 - √15] = 2√15Therefore, the distance is sqrt[(Δx)^2 + (Δy)^2] = sqrt[(√15)^2 + (2√15)^2] = sqrt[15 + 4*15] = sqrt[15 + 60] = sqrt[75] = 5√3Wait, that seems straightforward. Let me check if that makes sense.Alternatively, maybe there's a quicker way. Since the chord of a parabola can sometimes be found using the formula related to the roots. Let me think.Alternatively, when we have a quadratic equation in x, the difference between the roots is sqrt(D)/a, but here we need the actual distance between the two points.But seems like my calculation is correct. The difference in x is √15, difference in y is 2√15, so the distance squared is 15 + (4*15) = 15 + 60 = 75, sqrt(75) is 5√3. Yeah, 75 is 25*3. So 5√3. So the length is 5√3. That should be the answer for part (1).Now part (2): If the line has only one intersection point with the parabola, determine the possible value(s) of the slope k.So, the line is passing through (0, -1) with slope k. So its equation is y = kx -1. Wait, similar to part (1). For the line to have only one intersection point with the parabola, it must be tangent to the parabola. So, the condition is that the system of equations y² = 12x and y = kx -1 has exactly one solution. So substituting, we can get a quadratic equation and set its discriminant to zero.Let me do that.Substitute y = kx -1 into y² = 12x:(kx -1)^2 = 12xExpand left side: k²x² - 2kx +1 = 12xBring all terms to one side:k²x² - (2k +12)x +1 =0This is a quadratic in x. For there to be exactly one solution, discriminant must be zero.Compute discriminant D:D = [-(2k +12)]² - 4*k²*1 = (2k +12)^2 -4k²Expand (2k +12)^2: 4k² +48k +144So D = 4k² +48k +144 -4k² = 48k +144Set D =0:48k +144 =0Solve for k:48k = -144 => k = -144/48 = -3Therefore, the only possible value is k = -3.Wait, that seems straightforward. Let me check again.So substituting the line into parabola:(kx -1)^2 =12xk²x² -2kx +1 =12xk²x² - (2k +12)x +1 =0Discriminant D = [-(2k +12)]² -4*k²*1 = (2k+12)^2 -4k²Which is 4k² +48k +144 -4k² =48k +144Set equal to zero: 48k +144=0 => k= -3Therefore, the slope k must be -3. So the answer is k= -3.Hmm. Let me confirm with another approach. Alternatively, the condition for a line to be tangent to a parabola can be derived using calculus. For the parabola y²=12x, we can differentiate implicitly to find the slope of tangent at a point.Differentiate both sides:2y dy/dx =12 => dy/dx =6/ySuppose the tangent line at point (x1, y1) on the parabola has slope 6/y1. The equation of tangent line can be written as:yy1 =6(x +x1)But since the tangent line must pass through (0, -1), substitute x=0, y=-1:(-1)y1 =6(0 +x1) => -y1 =6x1But the point (x1,y1) is on the parabola y1²=12x1. So we have two equations:-y1=6x1y1²=12x1From first equation: x1= -y1/6Substitute into second equation:y1²=12*(-y1/6)= -2y1Therefore:y1² +2y1=0 => y1(y1 +2)=0So y1=0 or y1= -2If y1=0, then x1=0, but the tangent at (0,0) is x=0, which is the axis, but the line x=0 is vertical, slope undefined, but our line has slope k, which is defined. So y1=0 gives a vertical tangent, which is not relevant here since k is defined. So discard y1=0.If y1= -2, then x1= -(-2)/6= 2/6=1/3Therefore, the tangent line at (1/3, -2) is:y*(-2)=6(x +1/3)Simplify:-2y =6x +2Divide both sides by -2:y= -3x -1Thus, the equation of the tangent line is y= -3x -1, which has slope k=-3. Therefore, confirming that k=-3 is the only value. Therefore, part (2) answer is k=-3.So, both parts make sense. Part (1) length is 5√3, part (2) k=-3.Final Answer(1) The length of the chord is boxed{5sqrt{3}}.(2) The possible value of the slope ( k ) is boxed{-3}.</think>Given the parabola ( y^2 = 12x ) and a straight line with slope ( k ) passing through the point ( (0, -1) ):1. When the slope ( k = 2 ): - The equation of the line is ( y = 2x - 1 ). - Substitute ( y = 2x - 1 ) into ( y^2 = 12x ): [ (2x - 1)^2 = 12x ] - Expand and simplify: [ 4x^2 - 4x + 1 = 12x implies 4x^2 - 16x + 1 = 0 ] - Solve the quadratic equation using the quadratic formula: [ x = frac{16 pm sqrt{256 - 16}}{8} = frac{16 pm 4sqrt{15}}{8} = 2 pm frac{sqrt{15}}{2} ] - Find the corresponding ( y )-coordinates: [ y = 2 left(2 + frac{sqrt{15}}{2}right) - 1 = 3 + sqrt{15} quad text{and} quad y = 2 left(2 - frac{sqrt{15}}{2}right) - 1 = 3 - sqrt{15} ] - Calculate the distance between the points ( left(2 + frac{sqrt{15}}{2}, 3 + sqrt{15}right) ) and ( left(2 - frac{sqrt{15}}{2}, 3 - sqrt{15}right) ): [ text{Distance} = sqrt{(sqrt{15})^2 + (2sqrt{15})^2} = sqrt{15 + 60} = sqrt{75} = 5sqrt{3} ]2. When the line has only one intersection point with the parabola: - The equation of the line is ( y = kx - 1 ). - Substitute ( y = kx - 1 ) into ( y^2 = 12x ): [ (kx - 1)^2 = 12x ] - Expand and simplify: [ k^2x^2 - 2kx + 1 = 12x implies k^2x^2 - (2k + 12)x + 1 = 0 ] - Set the discriminant to zero for a single solution: [ (2k + 12)^2 - 4k^2 = 0 implies 48k + 144 = 0 implies k = -3 ] Final Answer(1) The length of the chord is (boxed{5sqrt{3}}).(2) The possible value of the slope ( k ) is (boxed{-3}).]

🔑:(1) Let's denote the intersection points of the line and the parabola as A(x_1, y_1) and B(x_2, y_2).Since the slope of the line is k=2 and it passes through (0, -1), the equation of the line is given by y - (-1) = 2(x - 0), which simplifies to y = 2x - 1.Solving the system of equations:begin{cases}y^2 = 12x y = 2x - 1end{cases}we substitute y from the second equation into the first one to get 4x^2 - 16x + 1 = 0.Using the Vieta's formulas, we have x_1 + x_2 = 4 and x_1 x_2 = frac{1}{4}.The length of chord AB can be found using the distance formula: |AB| = sqrt{1+k^2} sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = boxed{5sqrt{3}}.(2) For a general slope k, the equation of the line becomes y - (-1) = kx, which simplifies to y = kx - 1.Solving the system of equations:begin{cases}y^2 = 12x y = kx - 1end{cases}we substitute y from the second equation into the first one to get k^2 x^2 - (2k + 12)x + 1 = 0.For the line to intersect the parabola at only one point, the discriminant of the quadratic equation must be zero. Thus, setting the discriminant Delta = (2k + 12)^2 - 4k^2 equal to zero, we get: (2k + 12)^2 - 4k^2 = 0. Solving for k, we find k = 0 or k = -3.When k = 0, the line y = -1 intersects the parabola at the point left(frac{1}{12}, -1right).When k = -3, the line y = -3x - 1 is tangent to the parabola, hence also intersecting at only one point.In conclusion, the slope k must be boxed{0} or boxed{-3} for the line to have only one intersection point with the parabola.