Question

In Fig. 6.43, if $PQ\perp PS,PQ\parallel SR,\angle SQR={28}^0$ and $\angle QRT={65}^0$, then find the values of $x$ and $y$.

Answer 1

Given, $PQ\perp PS,PQ\parallel SR,\angle SQR={28}^{\circ },\angle QRT={65}^{\circ }$
According to the question,
$x+\angle SQR=\angle QRT$ (Alternate angles as $QR$ is transversal.)
$\Rightarrow x+{28}^{\circ }={65}^{\circ }$
$\Rightarrow x={37}^{\circ }$
Also $\angle QSR=x$
$\Rightarrow \angle QSR={37}^{\circ }$
Also $\angle QRS+\angle QRT={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle QRS+{65}^{\circ }={180}^{\circ }$
$\Rightarrow \angle QRS={115}^{\circ }$
Now, $\angle P+\angle Q+\angle R+\angle S={360}^{\circ }$ (Sum of the angles in a quadrilateral.)
$\Rightarrow {90}^{\circ }+{65}^{\circ }+{115}^{\circ }+\angle S={360}^{\circ }$

$\Rightarrow {270}^{\circ }+y+\angle QSR={360}^{\circ }$

$\Rightarrow {270}^{\circ }+y+{37}^{\circ }={360}^{\circ }$

$\Rightarrow {307}^{\circ }+y={360}^{\circ }$
$\Rightarrow y={53}^{\circ }$