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Question
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.
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°
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