JEE Main · 2005 · 08 May 2005 · Offline · Chapter_pending

A B C

is a triangle. Forces

P, Q, R

acting along

I A

I B

and

I C

respectively are in equilibrium, where

I

is the incentre of

△ A B C

. Then

P : Q : R

Answer: (C)

Solution

The forces act along

I A

I B

and

I C

, and the angle between the directions

I B

and

I C

∠ B I C = \frac{π}{2} + \frac{A}{2}

; likewise

∠ C I A = \frac{π}{2} + \frac{B}{2}

and

∠ A I B = \frac{π}{2} + \frac{C}{2}

. By Lami's theorem, each force is proportional to the sine of the angle between the other two, so

P : Q : R = sin (\frac{π}{2} + \frac{A}{2}) : sin (\frac{π}{2} + \frac{B}{2}) : sin (\frac{π}{2} + \frac{C}{2})

. Since

sin (\frac{π}{2} + θ) = cos θ

, this ratio is

cos \frac{A}{2} : cos \frac{B}{2} : cos \frac{C}{2}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option