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 |
\[\angle RQP=180{}^\circ -\left( \frac{\angle C}{2}+\angle B+\frac{\angle A}{2} \right)\] |
\[=180{}^\circ -\left( \frac{\angle A+\angle B+\angle C+\angle B}{2} \right)\] |
\[=180{}^\circ -\left( \frac{180{}^\circ +\angle B}{2} \right)\] |
\[=180{}^\circ -90{}^\circ -\frac{\angle B}{2}=90{}^\circ -\frac{\angle B}{2}\] |
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