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 A 2 sin B 2 sin C 2 ) is equal to _________ . [2023]

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Solution

Q.
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 _________ . [2023]

Ans.
(5)

$Any point on the given line \frac{x}{1} = \frac{y}{1} = \frac{z}{1} = λ is (λ, λ, λ)$

$If it intersects the given lines then it must satisfy them.$

$\Rightarrow λ (\sin A + \sin B + \sin C) = 2 \times 3^{2}$ ...(i)

$and λ (\sin 2 A + \sin 2 B + \sin 2 C) = 3^{2}$ ...(ii)

$On dividing equation (ii) by (i), we get$

$\frac{\sin 2 A + \sin 2 B + \sin 2 C}{\sin A + \sin B + \sin C} = \frac{1}{2}$ $\Rightarrow \frac{4 \sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$

$\Rightarrow 8 [\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}] = \frac{1}{2} \Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$

$∴$ $80 (\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 80 \times \frac{1}{16} = 5$

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