Question

In the given figure, PQ is the diameter of a circle with centre O. If $\angle PQR={65}^o,\angle SPR={40}^o$ and $\angle PQM={50}^o$. Find (i) $\angle QPR$ (ii) $\angle QPM$ and (iii) $\angle PRS.$

Answer 1

$\left(1\right)$ $\angle QPR={180}^{\circ }\angle PRQ-\angle RQP$

$\angle QPR={180}^{\circ }-{90}^{\circ }-{65}^{\circ }={25}^{\circ }$

$\left(2\right)$ $\angle QPM={180}^{\circ }\angle PMQ-\angle PQM$

$\angle QPM={180}^{\circ }-{90}^{\circ }-{50}^{\circ }={40}^{\circ }$

$\left(3\right)$ $\angle PSR={180}^{\circ }-\angle PQR={180}^{\circ }-{65}^{\circ }={115}^{\circ }$

$\angle PRS={180}^{\circ }\angle SPR-\angle PSR$

$\angle PRS={180}^{\circ }-{40}^{\circ }-{11}^{\circ }={15}^{\circ }$