Given:

∠APB = 50°

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a triangle = 180°.

By property 1,

AP = BP (tangent from P)

Therefore, ∠PAB = ∠PBA

Now,

By property 3 in ∆PAB,

∠PAB + ∠PBA + ∠APB = 180°

⇒ ∠PAB + ∠PBA = 180° – ∠APB

⇒ ∠PAB + ∠PBA = 180° – 50°

⇒ ∠PAB + ∠PBA = 130°

By property 2,

∠PAO = 90°

Now,

∠PAO = ∠PAB + ∠OAB

⇒ ∠OAB = ∠PAO – ∠PAB

⇒ ∠OAB = 90° – 65° = 25°

Hence, ∠OAB = 25°