In the given figure, O is the center of the circle, PA is tangent and PBC is secant. If ∠ABC = 60°; ∠P is :

1. 30°

2. 60°

3. 120°

4. 90°

In △ABC,

∠BAC = 90° (Angle in semi-circle is a right angle)

⇒ ∠ABC + ∠BAC + ∠ACB = 180° (By angle sum property of triangle)

⇒ 60° + 90° + ∠ACB = 180°

⇒ 150° + ∠ACB = 180°

⇒ ∠ACB = 180° - 150° = 30°.

We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

⇒ ∠BAP = ∠ACB = 30°.

From figure,

⇒ ∠PBA + ∠ABC = 180° [Linear pairs]

⇒ ∠PBA + 60° = 180°

⇒ ∠PBA = 180° - 60° = 120°.

In △PBA,

⇒ ∠PBA + ∠BAP + ∠APB = 180° (By angle sum property of triangle)

⇒ 120° + 30° + ∠APB = 180°

⇒ 150° + ∠APB = 180°

⇒ ∠APB = 180° - 150° = 30°.

∴ ∠P = 30°.

Hence, Option 1 is the correct option.