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The line x = y = z (parametric: x=t, y=t, z=t) intersects the pair of planes x*sinA + y*sinB + z*sinC = 18 and x*sin2A + y*sin2B + z*sin2C = 9, where A, B, C are angles of a triangle. Find the value of 80*(sinA/2 * sinB/2 * sinC/2).

1. 5
2. 10
3. 2.5
4. 20

Correct answer: 5

Solution

Substituting x=y=z=t: t*(sinA+sinB+sinC)=18...(1) and t*(sin2A+sin2B+sin2C)=9...(2). For triangle angles: sinA+sinB+sinC = 4*cos(A/2)*cos(B/2)*cos(C/2) and sin2A+sin2B+sin2C = 4*sinA*sinB*sinC. Divide (2) by (1): (sin2A+sin2B+sin2C)/(sinA+sinB+sinC) = 9/18 = 1/2. So 4*sinA*sinB*sinC / (4*cos(A/2)*cos(B/2)*cos(C/2)) = 1/2. Now sinA*sinB*sinC = 8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2). So [8*sin(A/2)*cos(A/2)*sin(B/2)*cos(B/2)*sin(C/2)*cos(C/2)] / [cos(A/2)*cos(B/2)*cos(C/2)] = 1/2. This gives 8*sin(A/2)*sin(B/2)*sin(C/2) = 1/2. So sin(A/2)*sin(B/2)*sin(C/2) = 1/16. Therefore 80*(sinA/2 * sinB/2 * sinC/2) = 80/16 = 5.

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