waecmaths question:
In the diagram $\left| SP \right|=\left| SR \right|$ and $\angle PRS={{50}^{\circ }}$ Calculate $\angle PQR$
Option A:
120o
Option B:
110o
Option C:
100o
Option D:
80o
waecmaths solution:
$\begin{align} & \angle RPS=\angle PRS={{50}^{\circ }}\text{ }\!\!\{\!\!\text{ Base }\angle s\text{ of Isso }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle RSP={{180}^{\circ }}-\angle RPS-\angle PRS\text{ }\!\!\{\!\!\text{ }\angle s\text{ in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & \angle RSP={{180}^{\circ }}-{{50}^{\circ }}-{{50}^{\circ }}={{80}^{\circ }} \\ & \angle RSP+\angle PQR={{180}^{\circ }}\text{ }\!\!\{\!\!\text{ opp }\angle s\text{ cyclic quad }\!\!\}\!\!\text{ } \\ & \text{8}{{\text{0}}^{\circ }}+\angle PQR={{180}^{\circ }} \\ & \angle PQR={{100}^{\circ }} \\\end{align}$
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