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Question
From a point P, two tangents PA and PB are drawn to a circle C(0, r). If OP = 2r, then find ∠APB. What type of triangle is APB?
Solution
Let ∠APO = θ
Sinθ = `(OA)/(OP) = 1/2` ⇒ θ = 30°
⇒ ∠APB = 2θ = 60°
Also ∠PAB = ∠PBA = 60° ......(∵ PA = PB)
⇒ ΔAPB is equilateral
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