In the diagram \[\left| PR \right|=\left| QR \right|\] and $\left| PR \right|=\left| RS \right|=\left| SP \right|$ , calculate the size of $\angle QRS$
150o
120o
90o
60o
\[\begin{align} & \text{In }\vartriangle PRS \\ & y+y+y={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ }\angle \text{s in a equilateral }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & y={{60}^{\circ }} \\ & \angle PSR+\angle PQR={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ opp }\angle s\text{ of a cyclic quad }\!\!\}\!\!\text{ } \\ & \text{6}{{\text{0}}^{\circ }}+\angle PQR={{180}^{\circ }} \\ & \angle PQR={{120}^{\circ }} \\ & In\text{ }\vartriangle PQR \\ & \angle QPR=\angle QRP\text{ }\!\!\{\!\!\text{ Base }\angle s\text{ of Iss}\text{. }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & x+x+{{120}^{\circ }}={{180}^{\circ }} \\ & x={{30}^{\circ }} \\ & \angle QRS=x+y={{30}^{\circ }}+{{60}^{\circ }}={{90}^{\circ }} \\\end{align}\]
