If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that $∠$ QPT = $70^{o}$ , then find the measure of $∠$ POQ.

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Solution

We know that the radius and tangent are perpendicular at their point of contact.

<OPT = 90°

Now, <OPQ = <OPT – <TPQ = 90° – 70° = 20°

Since OP = OQ as both are radius

<OPQ = <OQP = 20° (Angles opposite to equal sides are equal)

Now, In isosceles <POQ

<POQ + <OPQ + <OQP = 180° (Angle sum property of a triangle)

<POQ = 180° – 20° – 20° = 140°

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