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Three forces $\vec{P}, \vec{Q}$ and $\vec{R}$ acting along IA, IB and IC, where I is the incentre of a $\triangle A B C$, are in equilibrium. Then $\vec{P}: \vec{Q}: \vec{R}$ is
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Verified Answer
The correct answer is:
$\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$
$\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$
By Lami's theorem
$$
\begin{aligned}
& \vec{P}: \vec{Q}: \vec{R}=\sin \left(90^{\circ}+\frac{A}{2}\right): \sin \left(90^{\circ}+\frac{B}{2}\right): \sin \left(90^{\circ}+\frac{C}{2}\right) \\
& \Rightarrow \cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2} .
\end{aligned}
$$
$$
\begin{aligned}
& \vec{P}: \vec{Q}: \vec{R}=\sin \left(90^{\circ}+\frac{A}{2}\right): \sin \left(90^{\circ}+\frac{B}{2}\right): \sin \left(90^{\circ}+\frac{C}{2}\right) \\
& \Rightarrow \cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2} .
\end{aligned}
$$
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