# 1. In Fig. 6.39, sides QP and RQ of $\Delta$PQR are produced to points S and T respectively. If $\angle$ SPR = 135° and $\angle$ PQT = 110°, find $\angle$ PRQ.

Given,
PQR is a triangle, $\angle$SPR =$135^0$,  $\angle$PQT = $110^0$

Now, $\angle$TQP + $\angle$PQR = $180^0$ (Linear pair)
So, $\angle$PQR = $180^0-110^0 = 70^0$

Since the side of QP of the triangle, PQR is produced to S
So, $\angle$PQR + $\angle$PRQ = $135^0$ (Exterior angle property of triangle)
$\therefore \angle PRQ = 135^0 - 70^0 = 65^0$

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