If PT is a tangent to a circle with center O and PQ is a chord of the circle such that ∠QPT = 70°, 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 Δ POQ

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

⇒ ∠POQ =180° - 20° = 140°

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

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