In Fig. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95°and TSQ = 75°, find SQT.

It is given in the question that:

PRT = 40o


RPT = 95o and,


TSQ = 75o


Now according to the question,


PRT + RPT + PTR = 180o (Sum of interior angles of the triangle)


40o + 95o + PTR = 180o


40o + 95o + PTR = 180o


135o + PTR = 180o


PTR = 45o


PTR = STQ = 45o (Vertically opposite angles)


Now,


TSQ + PTR + SQT = 180o (Sum of the interior angles of the triangle)


75o + 45o + SQT = 180o


120o + SQT = 180o


SQT = 60o


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