Mathematics
Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that :
∠PAQ = 2∠OPQ
Circles
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Answer
We know that,
The tangent at any point of a circle and the radius through this point are perpendicular to each other.
In quadrilateral OPAQ,
∠OPA = ∠OQA = 90°.
∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360° [∵ Sum of angles in quadrilateral = 360°]
90° + 90° + ∠POQ + ∠PAQ = 360°
∠POQ + ∠PAQ = 360° - 180°
∠POQ + ∠PAQ = 180° ……….(1)
In △OPQ,
OP = OQ [Radius of same circle]
∴ ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]
By angle sum property of triangle,
⇒ ∠OPQ + ∠OQP + ∠POQ = 180°
⇒ ∠OPQ + ∠OPQ + ∠POQ = 180°
⇒ 2∠OPQ + ∠POQ = 180° ………(2)
From (1) and (2) we get,
∠POQ + ∠PAQ = 2∠OPQ + ∠POQ
⇒ ∠PAQ = 2∠OPQ.
Hence, proved that ∠PAQ = 2∠OPQ.
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