question_answer

Three forces P, Q and R acting along\[IA,IB\]and \[IC,\]where\[I\]is the incentre of a\[\Delta ABC,\]are in equilibrium. Then, P : Q : R is

A) \[\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \frac{C}{2}\]

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) \[\cos ec\,\frac{A}{2}:\,\,\cos ec\,\frac{B}{2}\,:\,\cos ec\,\frac{C}{2}\]

Correct Answer: A

Solution :

Three forces P, Q and R acting along\[lA,lB\]and \[lC\]are in equilibrium. \[\angle AlB=\pi -\frac{\angle A+\angle B}{2}=\pi -\left( \frac{\pi }{2}-\frac{C}{2} \right)=\frac{\pi }{2}+\frac{C}{2}\] Similarly, \[\angle BlC=\frac{\pi }{2}+\frac{A}{2}\] and     \[\angle AlC=\frac{\pi }{2}+\frac{B}{2}\] By Lami's theorem, \[\frac{P}{\sin \angle BlC}=\frac{Q}{\sin \angle AlC}=\frac{R}{\sin \angle AlB}\] \[\Rightarrow \]\[\frac{P}{\sin \left( \frac{\pi }{2}+\frac{A}{2} \right)}=\frac{Q}{\sin \left( \frac{\pi }{2}+\frac{B}{2} \right)}=\frac{R}{\sin \left( \frac{\pi }{2}+\frac{C}{2} \right)}\] \[\Rightarrow \]\[\frac{P}{\cos \frac{A}{2}}=\frac{Q}{\cos \frac{B}{2}}=\frac{R}{\cos \frac{C}{2}}\] \[\Rightarrow \]\[p=\lambda \cos A/2,q=\lambda \cos B/2]\] \[\therefore \]\[p:q:r=\cos \frac{A}{2}:\cos \frac{B}{2}:\cos \frac{C}{2}\]