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?

Answer 1

Let ∠APO = θ

Sinθ = `(OA)/(OP) = 1/2` ⇒ θ = 30°

⇒ ∠APB = 2θ = 60°

Also ∠PAB = ∠PBA = 60° ......(∵ PA = PB)

⇒ ΔAPB is equilateral