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


It is given in the question that:

SPR = 135O


And,


PQT = 110o


Now, according to the question,


SPR + QPR = 180O (SQ is a straight line)


135o + QPR = 180O


QPR = 45O


And,


PQT + PQR = 180O (TR is a straight line)


110o + PQR = 180O


PQR = 70O


Now,


PQR + QPR + PRQ = 180O (Sum of the interior angles of the triangle)


70o + 45o + PRQ = 180O


115O + PRQ = 180O


PRQ = 65O


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