In the figure,if PQ perp PS,PQ parallel SR,an | vedclass

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Question

(A) In $\Delta QRS$,the side $SR$ is produced to $T$.Therefore,the exterior angle $\angle QRT = \angle RQS + \angle RSQ$ (Exterior angle property).Giv

Vedclass · Accepted Answer

$x = 37^o$ and $y = 53^o$

Vedclass · Answer

(A) In $\Delta QRS$,the side $SR$ is produced to $T$.Therefore,the exterior angle $\angle QRT = \angle RQS + \angle RSQ$ (Exterior angle property).Given $\angle RQS = 28^o$ and $\angle QRT = 65^o$.So,$28^o + \angle RSQ = 65^o$.$\angle RSQ = 65^o - 28^o = 37^o$.Since $PQ \parallel SR$ and $QS$ is a transversal,the alternate interior angles are equal.Therefore,$\angle PQS = \angle RSQ$.$x = 37^o$.Given $PQ \perp PS$,so $\angle SPQ = 90^o$.In $\Delta PQS$,by the angle sum property,$\angle SPQ + \angle PQS + \angle PSQ = 180^o$.$90^o + x + y = 180^o$.$90^o + 37^o + y = 180^o$.$127^o + y = 180^o$.$y = 180^o - 127^o = 53^o$.Thus,$x = 37^o$ and $y = 53^o$.

In the figure,if $PQ \perp PS$,$PQ \parallel SR$,$\angle SQR = 28^o$ and $\angle QRT = 65^o$,then find the values of $x$ and $y$.

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