Q3 of 47 Page 1

In Fig. 1, O is the centre of the circle, PQ is a chord and PT is tangent to the circle at P. If POQ = 70°, find TPQ.


(Fig. 1)

Given: POQ = 70°

PO = OQ (Represents the radius of the same circle)


So ΔOPQ is isosceles


OPQ = OQP (Base angles of an isosceles triangle) …Equation (i)


OPQ + OQP + POQ = 180°(Sum of interior angles of a triangle)


⇒ OPQ + OQP = 180°-70° = 110°


⇒ 2 × OPQ = 110° (Equation (i))


⇒ OPQ = 55°


OPT = 90° (Tangents are always perpendicular with the line joining the center)


TPQ = OPT-OPQ


⇒ TPQ = 90°-55°


Answer: TPQ = 35°


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