Question

$ABC$ is a triangle. Forces $\vec{P},\vec{Q},\vec{R}$ acting along $IA,IB$ and $IC$ respectively are in equilibrium, where $I$ is the in centre of $\Delta ABC$. Then $\vec{P}:\vec{Q}:\vec{R}$ is :

• A. $cosA:cosB:cosC$
• B. $cos\frac{A}{2}:cos\frac{B}{2}:cos\frac{C}{2}$
• C. $sin\frac{A}{2}:sin\frac{B}{2}:sin\frac{C}{2}$
• D. $sinA:sinB:sinC$

Answer 1

Answer: (B)

If three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two.
$\therefore \frac{P}{\sin \alpha }=\frac{Q}{\sin \beta }=\frac{R}{\sin \gamma }$
This theorem is known as Lami’s theorem.
In a $\Delta ABC,I$ is the in centre.
$\therefore \angle BIC$$=\pi -\left(\frac{B}{2}+\frac{C}{2}\right)$
$=\pi -\left(\frac{\pi }{2}-\frac{A}{2}\right)$
$=\frac{\pi }{2}+\frac{A}{2}$
Similarly, $\angle AIC$ $\frac{\pi }{2}+\frac{B}{2}$ and $\angle AIB\frac{\pi }{2}+\frac{C}{2}$
By Lami’s theorem
$\frac{\vec{A}}{\sin BIC}=\frac{\vec{B}}{\sin AIC}\frac{\vec{R}}{\sin AIB}$
$\therefore \vec{P}:\vec{Q}:\vec{R}$
$=\sin \left(\frac{\pi }{2}+\frac{A}{2}\right):\left(\frac{\pi }{2}+\frac{B}{2}\right):\sin \left(\frac{\pi }{2}+\frac{C}{2}\right)$
$=\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \frac{C}{2}$