Question

If the line x = y = z intersect the line

sinA. x + sinB.y + sinC.z = 2d², sin2A.x + sin 2B.y + sin2C.z = d², then sin . sin . sin is equal to

(where A, B, C are the angles of a triangle) –

• A. $\frac { 1 } { 16 }$
• B. $\frac { 1 } { 8 }$
• C. $\frac { 1 } { 32 }$
• D. $\frac { 1 } { 12 }$

Answer 1

Answer: (A)

$\frac { 1 } { 16 }$

Answer 2

Let the point of intersection be (l,l,l)

\ sinA + sinB + sin C = 2 (sin2A + sin 2B + sin2C)

\ 4cos cos cos = 8 sin A sin B sin C

\ sin sin sin $\frac { \mathrm { C } } { 2 }$ = $\frac { 1 } { 16 }$.