question_answer

ABC is a cyclic triangle and the bisectors of \[\angle BAC,\]\[\angle ABC\] and \[\angle BCA\] meet the circle at P, Q and R, respectively. Then, the \[\angle RQP\]is                                                                              [SSC (CGL) Pre 2015]

A) \[90{}^\circ -\,\,\frac{\angle B}{2}\]       

B) \[90{}^\circ +\frac{\angle C}{2}\]

C) \[90{}^\circ -\,\,\frac{\angle A}{2}\]                   

D) \[90{}^\circ +\frac{\angle B}{2}\]

Correct Answer: A

Solution :

Here, RBPQ makes a cyclic quadrilateral.

\[\therefore \]      \[\angle RQP+\angle RBP=180{}^\circ \]

\[\angle RQP=180{}^\circ -\angle RBP\]              (i)

Now, \[\angle RBP=\angle RBA+\angle B+\angle CBP\]

\[\angle RBP=\frac{\sqrt{C}}{2}+\angle B+\frac{\angle A}{2}\]    (ii)

Put value of \[\angle RBP\]in Eq. (i), we get

\[\angle RQP=180{}^\circ -\left( \frac{\angle C}{2}+\angle B+\frac{\angle A}{2} \right)\]

\[=180{}^\circ -\left( \frac{\angle A+\angle B+\angle C+\angle B}{2} \right)\]

\[=180{}^\circ -\left( \frac{180{}^\circ +\angle B}{2} \right)\]

\[=180{}^\circ -90{}^\circ -\frac{\angle B}{2}=90{}^\circ -\frac{\angle B}{2}\]