Question

Three forces $\overline{P},\overline{Q},\overline{R}$ acting along $I_A,I_B,I_C$ where $I$ is the incentre of a triangle, are in equilibrium then $\overline{P}:\overline{Q}:\overline{R}$

• A. $\sec \frac{A}{2}:\sec \frac{B}{2}:\sec \frac{C}{2}$
• B. $\sin \frac{A}{2}:\sin \frac{B}{2}:\sin \frac{C}{2}$
• C. $\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \frac{C}{2}$
• D. $\text{cosec}\frac{A}{2}:\text{cosec}\frac{B}{2}:\text{cosec}\frac{C}{2}$

Answer 1

Answer: (C)

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

$\angle AIC=\pi -\frac{A+C}{2}$
$\angle AIC=\pi -(\frac{\pi }{2}-\frac{B}{2})$
$\frac{\pi }{2}+\frac{B}{2},$

Similarly $\angle BIC=\frac{\pi }{2}+\frac{A}{2}$ and $\angle AIB=\frac{\pi }{2}+\frac{C}{2}$

Now using Lami's theorem we have

$\frac{\overrightarrow{p}}{\sin (\frac{\pi }{2}+\frac{A}{2})}$=$\frac{\overrightarrow{Q}}{\sin (\frac{\pi }{2}+\frac{B}{2})}=\frac{\overrightarrow{R}}{\sin (\frac{\pi }{2}+\frac{C}{2})}$

Hence $\overrightarrow{P}:\overrightarrow{Q}:\overrightarrow{R}$ is $\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \frac{C}{2}$