In the diagram, TS is a tangent to the circle at S . $\left| PR \right|=\left| RS \right|$ and $\angle PQR={{117}^{\circ }}$. Calculate $\angle PST$
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$\begin{align} & \angle PSR={{180}^{\circ }}-\angle PQR\text{ }\!\!\{\!\!\text{ sum of opp}\text{. }\angle s\text{ in cyclic quad }\!\!\}\!\!\text{ } \\ & \angle PSR={{180}^{\circ }}-{{117}^{\circ }}={{63}^{\circ }} \\ & \angle RPS=\angle PSR={{63}^{\circ }}\text{ }\!\!\{\!\!\text{ Base }\angle s\text{ of Isso}\text{. }\vartriangle \} \\ & \angle RPS+\angle PSR+\angle PRS={{180}^{\circ }}^{\circ }\text{ }\!\!\{\!\!\text{ sum of }\angle s\text{ in }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \text{6}{{\text{3}}^{\circ }}+{{63}^{\circ }}+\angle PRS={{180}^{\circ }} \\ & \angle PRS={{180}^{\circ }}-{{126}^{\circ }}={{54}^{\circ }} \\ & \angle PST=\angle PRS={{54}^{\circ }}\text{ }\!\!\{\!\!\text{ Angles in opposite segment }\!\!\}\!\!\text{ } \\\end{align}$
