question_answer

ABC is a triangle. Forces P,Q,R acting along \[I,A,IB\]and\[IC\]respectively are in equilibrium, where I is the incentre of\[\Delta ABC.\]Then, P : Q : R is     AIEEE  Solved  Paper-2005

A) \[\cos A:\cos B:\cos \,C\]

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) \[\sin A:\sin B:\sin C\]

Correct Answer: B

Solution :

If three forces acting on a particle keep it in equilibrium, each is proportional to the sine of the angle between the other two, then This theorem is known as Lami?s theorem. In \[\Delta ABC,\,\,l\] is the in centre. \[\angle BlC=\pi -\left( \frac{B}{2}+\frac{C}{2} \right)\] \[=\pi -\left( \frac{\pi }{2}-\frac{A}{2} \right)=\frac{\pi }{2}\,+\frac{A}{2}\] Similarly \[\angle AlC\,=\frac{\pi }{2}\,+\frac{B}{2}\] and \[\angle AlB\,=\frac{\pi }{2}+\frac{C}{2}\] By Lami's theorem, \[\frac{P}{\sin \,BlC}=\frac{Q}{\sin AlC}\,=\frac{R}{\sin AlB}\]