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

                

Answers (1)


Given,
PQR is a triangle, \angleSPR =135^0,  \anglePQT = 110^0

Now, \angleTQP + \anglePQR = 180^0 (Linear pair)
So, \anglePQR = 180^0-110^0 = 70^0

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

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