In the given figure, PA and PB are two tangents from common point P

∴ PA = PB

[Tangents drawn from an external point are equal]

∠PBA = ∠PAB

[Angles opposite to equal angles are equal] [1]

By angle sum property of triangle in △APB

∠APB + ∠PBA + ∠PAB = 180°

50° + ∠PAB + ∠PAB = 180° [From 1]

2∠PAB = 130°

∠PAB = 65° [2]

Now,

∠OAP = 90°

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAB + ∠PAB = 90°

∠OAB + 65° = 90° [From 2]

∠OAB = 25°