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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