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In Figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.

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In Figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.

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It is given the TQR is a straight line and so, the linear pairs (i.e. TQP and PQR) will add up to 180°

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°

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