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10/07/2021 12:16 pm
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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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10/07/2021 12:21 pm
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°