In the diagram above, $\angle RPS={{50}^{o}}$, $\angle RPQ={{30}^{o}}$and PQ = QR . Find the value of $\angle PRS$
50o
60o
70o
80o
$\begin{align} & \angle QPR=\angle QRP={{30}^{o}}\text{ }\!\!\{\!\!\text{ Base angle of issosceles }\Delta \text{ }\!\!\}\!\!\text{ } \\ & \angle PQR+{{30}^{o}}+{{30}^{o}}={{180}^{o}}\text{ }\!\!\{\!\!\text{ Sum of }\angle \text{s in }\Delta PQR\} \\ & \angle PQR+\angle PSR={{180}^{o}}\text{ }\!\!\{\!\!\text{ sum of opp}\text{.}\angle \text{s in a cyclic quad }\!\!\}\!\!\text{ } \\ & \angle PSR={{180}^{o}}-{{120}^{o}}={{60}^{o}} \\ & \angle PSR+\angle PRS+\angle RPS=180\text{ }\!\!\{\!\!\text{ Sum of }\angle \text{s in }\Delta PSR\} \\ & {{60}^{o}}+\angle PRS+{{50}^{o}}={{180}^{o}} \\ & \angle PRS={{70}^{o}} \\\end{align}$