Question

In the given figure, PA and PB are tangents to a circle centred at O. If ∠AOB = 130°, then ∠APB is equal to:

Options

• 130°

• 50°

• 120°

• 90°

Answer 1

50°

Explanation:

Given:

PA and PB are tangents to a circle with centre O.

∠AOB = 130°

We know that the radius is perpendicular to the tangent at the point of contact.

∠OAP = 90°

∠OBP = 90°

In quadrilateral OAPB:

∠AOB + ∠OAP + ∠APB + ∠OBP = 360°

130° + 90° + ∠APB + 90° = 360°

310° + ∠APB = 360°

∠APB = 360° – 310°

∠APB = 50°