**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°