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

$PQ \perp PS, \angle SQR=28^o$ and $\angle QRT=65^o$.

To do:

We have to find the values of $x$ and $y$.

Solution:

Since,

$QR$ is a transversal, the alternate angles are added.

$x+\angle SQR=\angle QRT$

By substituting the values of $\angle QRT$ and $\angle SQR$ we get,

$x+28^o=65^o$

$x=65^o-28^o$

$x=37^o$

We also know that,

The lines intersected by the transversal are parallel, alternate interior angles are equal.

$\angle QSR=37^o$

We know that,

The sum of the measures of the angles in linear pairs is always $180^o$.

Therefore,

$\angle QRS+\angle QRT=180^o$

$\angle QRS+65^o=180^o$

This implies,

$\angle QRS=180^o-65^o$

$\angle QRS=115^o$

By using the angle sum property in $\triangle SPQ$

$\angle SPQ+x+y=180^o$

$90^o+37^o+y=180^o$

$y=180^o-127^o$

$y=53^o$

Hence, $y=53^o$.