In the given diagram, PA and PB are the tangents to the circle with centre O, such that PA = PB = AB. The value of ∠ACB is:

Given,

PA = PB = AB.

Since all three sides of ∆PAB are equal, ∆PAB is an equilateral triangle.

∴ ∠APB = 60°.

We know that,

Radius is perpendicular to the tangent at the point of contact.

∴ ∠OAP = ∠OBP = 90°.

In quadrilateral OAPB,

⇒ ∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

⇒ ∠AOB + 90° + 60° + 90° = 360°

⇒ ∠AOB = 360° - 240° = 120°.

We know that,

The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

$\Rightarrow \angle \text{ ACB }= \frac{1}{2} \angle \text{ AOB }= \frac{1}{2} \times 120° = 60°.$

Hence, option 3 is the correct option.