Question

In the isosceles triangle PQR, the vertical angle $\angle P=50°$. The circle passing through Q and R cuts PQ in S and PR in T. ST is joined. Find $\angle \mathrm{PST}$

Answer 1

$$\begin{gathered} \angle \mathrm{PQR}=\angle \mathrm{PRQ}(\because \mathrm{PQ}=\mathrm{PR}) \\ \mathrm{In}∆\mathrm{PQR},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{PQR}+\angle \mathrm{PRQ}+\angle \mathrm{QPR}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{PRQ}=180°-50° \\ \Rightarrow \angle \mathrm{PRQ}=65° \\ \angle \mathrm{PRQ}+\angle \mathrm{QST}=180°(\mathrm{STRQ}\mathrm{is}\mathrm{a}\mathrm{cylic}\mathrm{quadrilateral}) \\ \Rightarrow \angle \mathrm{QST}=180°-65°=115° \\ \angle \mathrm{PST}+\angle \mathrm{QST}=180°(\mathrm{Linnear}\mathrm{pair}) \\ \Rightarrow \angle \mathrm{PST}=180°-115°=65° \end{gathered}$$