Tangents PA and PB are drawn from an exterior point P. A and B lie on the circumference of the circle. If angle APB = 60°, the triangle PAB is :

1. isosceles

2. isosceles right angled

3. equilateral

4. right angled

We know that,

Tangents from an external point are equal in length.

∴ PA = PB

∴ ∠B = ∠A = x (let) (Angles opposite to equal sides are equal)

In △ APB,

⇒ ∠A + ∠B + ∠P = 180° (By angle sum property of triangle)

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

Since, ∠A = ∠B = ∠P = 60°

∴ △ APB is an equilateral triangle.

Hence, Option 3 is the correct option.