1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution: It is given the TQR is a straight line and so, the linear pairs (i.e. ∠TQP and ∠PQR) will add up to 180° So, ∠TQP +∠PQR = 180° Now, putting the value of ∠TQP = 110° we get, ∠PQR = 70° Consider the ΔPQR, Here, the side QP is extended to S and so, ∠SPR forms the exterior angle. Thus, ∠SPR (∠SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property) Or, ∠PQR +∠PRQ = 135° Now, putting the value of ∠PQR = 70° we get, ∠PRQ = 135°-70° Hence, ∠PRQ = 65°
