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°

60° + ∠PAB + ∠PAB = 180° [From 1]

2∠PAB = 120°

∠PAB = 60° …[2]

Now,

∠OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAB + ∠PAB = 90°

∠OAB + 60° = 90° [From 2]

∠OAB = 30°