Mathematics

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

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

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

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

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

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Subjective Medium Published on 09th 09, 2020
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