Question

In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If ∠PBA = 65°, find ∠OAB and ∠APВ.

Answer 1

Given:

PA and PB are tangents from external point P to a circle with centre O.

∠PBA = 65°.

Step-wise calculation:

1. Tangents from the same external point are equal: PA = PB.

Hence triangle PAB is isosceles, so base angles at A and B are equal: ∠PAB = ∠PBA = 65°.

2. Angle at P in triangle PAB:

∠APB = 180° – (∠PAB + ∠PBA)

= 180° – (65° + 65°)

= 50°

3. Radius to a point of contact is perpendicular to the tangent: OA ⟂ PA.

Therefore ∠OAP = 90°.

4. ∠OAB is the complement of ∠PAB (because ∠PAB + ∠OAB = 90°), so ∠OAB = 90° – 65° = 25°.