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ABC is a triangle. Forces P, Q, R acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of △ ABC. Then P: Q: R is

1. sin A: sin B: sin C
2. sin (A)/(2): sin (B)/(2): sin (C)/(2)
3. cos (A)/(2): cos (B)/(2): cos (C)/(2)
4. cos A: cos B: cos C

Correct answer: cos (A)/(2): cos (B)/(2): cos (C)/(2)

Solution

For forces along IA, IB, IC at the incentre I in equilibrium, each force is proportional to the cosine of half the opposite... actually each is proportional to cos of half its own vertex angle: P:Q:R = cos(A/2):cos(B/2):cos(C/2).

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