Question

In the given figure, PA and PB are tangents to a circle centred at O. If ∠OAB = 15°, then ∠APB equals:

Options

• 30°

• 15°

• 45°

• 10°

Answer 1

30°

Explanation:

Given:

∠OAB = 15°

In △OAB, OA = OB (radii of the same circle).

Since it is an isosceles triangle:

∠OBA = ∠OAB = 15°

In △OAB:

∠AOB + ∠OAB + ∠OBA = 180°

∠AOB + 15° + 15° = 180°

∠AOB = 180° – 30° = 150°

We know that the angle between two tangents from an external point is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

∠APB + ∠AOB = 180°

∠APB + 150° = 180°

∠APB = 30°