From the question we have:

*PA and PB are the tangents to the circle.
*

∴ OA ⊥ PA

⇒ ∠OAP = 90°

In ΔOPA,

sin ∠OPA = OA OP = r 2r [Given OP is the diameter of the circle]

⇒ sin ∠OPA = 1 2 = sin 30 ⁰

⇒ ∠OPA = 30°

*Similarly, it can be proved that ∠OPB = 30°.*

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB [length of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA ............(1) [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60° .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

**∴ ΔPAB is an equilateral triangle.**