If the line $\mathrm{x}=\mathrm{y}=\mathrm{z}$ intersects the line $\mathrm{x} \sin \mathrm{A}+\mathrm{y} \sin \mathrm{B}+\mathrm{z} \sin \mathrm{C}-18=0=\mathrm{x} \sin 2 \mathrm{~A}+\mathrm{y} \sin 2 \mathrm{~B}+\mathrm{z} \sin 2 \mathrm{C}-9$, where $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are the angles of a triangle ABC, then $80\left(\sin \frac{\mathrm{~A}}{2} \sin \frac{\mathrm{~B}}{2} \sin \frac{\mathrm{C}}{2}\right)$ is equal to ____

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Properties of Triangle

If the line

x = y = z

intersects the line

x sin A + y sin B + z sin C - 18 = 0 = x sin 2 A + y sin 2 B + z sin 2 C - 9

, where

A, B, C

are the angles of a triangle ABC, then

80 (sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2})

is equal to ____

Answer: 5

Since the line

x = y = z

meets both given surfaces, put

x = y = z = t

to get

t (sin A + sin B + sin C) = 18

and

t (sin 2 A + sin 2 B + sin 2 C) = 9

. Dividing,

\frac{sin A + sin B + sin C}{sin 2 A + sin 2 B + sin 2 C} = 2

. For a triangle,

sin A + sin B + sin C = 4 cos \frac{A}{2} cos \frac{B}{2} cos \frac{C}{2}

and

sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C

. Using

sin A = 2 sin \frac{A}{2} cos \frac{A}{2}

and likewise for

B, C

, the ratio reduces to

\frac{1}{8 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = 2

, so

16 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = 1

. Therefore

80 sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2} = \frac{80}{16} = 5

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer