Question

In the given figure if $PQ$ $\perp$ $PS,PQ||SR$, $\angle SQR={28}^o$ and $\angle QRT={65}^o$, then find the value of $x$ and $y$.

Answer 1

In $△QSR$, we have

$\angle QRT=\angle RSQ+\angle SQR$ ....(Exterior angle of triangle is equal t sum of interior opposite angles)

$\Rightarrow {65}^{\circ }=\angle RSQ+{28}^{\circ }$

$\Rightarrow \angle RSQ={65}^{\circ }-{28}^{\circ }$

$\Rightarrow \angle RSQ={37}^{\circ }$

We have $y^{\circ }+\angle RSQ={90}^{\circ }$ ....(Given)

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

$\Rightarrow y={90}^{\circ }-{37}^{\circ }={53}^{\circ }$

In $△PQS$, we have

$x^{\circ }+y^{\circ }+\angle P=180^{\circ }$ ....(Angle sum property)

$\Rightarrow x+{53}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow x={180}^{\circ }-{143}^{\circ }={37}^{\circ }$