Question

In fig, if $PQ\perp PS,PQ\parallel SR,\angle SQR={28}^0$ and $\angle QRT={65}^0$, then find the value of x and y.

Answer 1

Given $PQ\perp PS,PQ\parallel SR$

$\angle SQR={28}^{\circ },\angle QRT={65}^{\circ }$

$\angle QRS+\angle QRT={180}^{\circ }⟹\angle QRS={180}^{\circ }-{65}^{\circ }={115}^{\circ }$

$PQ\parallel SR⟹\angle PQS=\angle QSR=x$

In $△QRS,\angle QRS+\angle SQR+\angle QSR={180}^{\circ }$

$⟹x+{115}^{\circ }+{28}^{\circ }={180}^{\circ }⟹x={37}^{\circ }$

$\angle PSR={90}^{\circ }⟹x+y={90}^{\circ }⟹y={90}^{\circ }-{37}^{\circ }={63}^{\circ }$