Question

If $PQ\perp PS,PQ\parallel SR,\angle SOR={28}^o$ and $\angle QRT={65}^o$, then find $x$ & $y$.

Answer 1

$\because PQ\parallel SR$

$\angle PQR=\angle QRT$ (Alternate interior angle)

$\Rightarrow x+{28}^{\circ }={65}^{\circ }$

$\Rightarrow x={65}^{\circ }-{28}^{\circ }$

$\Rightarrow x={37}^{\circ }$

$In\Delta PQS$

$\angle P+\angle Q+\angle S={108}^{\circ }$ (By angle sum property)

$\Rightarrow {90}^{\circ }+x+y={180}^{\circ }$

$\Rightarrow {90}^{\circ }+{37}^{\circ }+y={180}^{\circ }$

$\Rightarrow y={108}^{\circ }-127$

$\Rightarrow y={53}^{\circ }$