Question

In the given figure $\square PTQR$ and PS is the bisector of $\angle P.$ If $\angle Q={60}^o$ and $\angle R={30}^o$, find $\angle TPS$.

Answer 1

In $△PQR,\angle P=180-(\angle Q+\angle R)=180-(60+30)={90}^{\circ }$

In $△QPT,\angle P=180-(\angle Q+\angle T)=180-(60+90)={30}^{\circ }$

In $△PTS,\angle S=180-(\angle T+\angle P)=180-(90+30)={60}^{\circ }$

$\therefore \angle TPS={30}^{\circ }$