# 5. In Fig. 6.43, if PQ $\bot$ PS, PQ $||$SR, $\angle$ SQR = 28° and $\angle$ QRT = 65°, then find the values of x and y.

M manish painkra

We have,
PQ $\bot$ PS, PQ || SR, $\angle$ SQR = 28° and $\angle$ QRT = 65°
Now, In $\Delta$ QRS, the side SR produced to T and PQ || RS
therefore, $\angle$QRT = $28^0+x$ = $65^0$
So, $x = 37^0$

Also, $\angle$QRT = $\angle$RSQ + $\angle$SQR (By exterior angle property of a triangle)
Therefore, $\angle$RSQ = $\angle$QRT - $\angle$SQR
$\angle RSQ = 65^0-28^0 = 37^0$

Now, in $\Delta$ PQS,
$\angle$P + $\angle$PQS + $\angle$PSQ = $180^0$
$y=180^0-90^0-37^0$
$y=52^0$

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