Three forces $\vec{P}, \vec{Q}$ and $\vec{R}$ acting along $I A$, $I B$ and $I C$, where $I$ is the incentre of a $\triangle A B C$, are in equilibrium. Then $\vec{P}: \vec{Q}: \vec{R}$ is

(A) $\cos \frac{\mathrm{A}}{2}: \cos \frac{\mathrm{B}}{2}: \cos \frac{\mathrm{C}}{2}$

(B) $\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}$

(C) $\sec \frac{A}{2}: \sec \frac{B}{2}: \sec \frac{C}{2}$

(D) $\operatorname{cosec} \frac{\mathrm{A}}{2}: \operatorname{cosec} \frac{\mathrm{B}}{2}: \operatorname{cosec} \frac{\mathrm{C}}{2}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Chapter_pending

Three forces

P, Q

and

R

acting along

I A

I B

and

I C

, where

I

is the incentre of a

△ A B C

, are in equilibrium. Then

P : Q : R

Answer: (A)

The three forces act outward from the incentre

I

along

I A

I B

and

I C

. A standard result gives the angles at the incentre as

∠ B I C = 9 0^{\circ} + \frac{A}{2}

∠ C I A = 9 0^{\circ} + \frac{B}{2}

and

∠ A I B = 9 0^{\circ} + \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 (9 0^{\circ} + \frac{A}{2}) : sin (9 0^{\circ} + \frac{B}{2}) : sin (9 0^{\circ} + \frac{C}{2}) = cos \frac{A}{2} : cos \frac{B}{2} : cos \frac{C}{2}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option