In the given figure, PA and PB are two tangents to the circle with centre O.

Construction: Join OB

We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360∘            [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90∘ + 50∘+ 90∘ = 360∘ 
⇒ 230∘+ ∠BOC = 360∘ 
⇒ ∠AOB = 130∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180∘            [Angle sum property of a triangle]
⇒ 130∘ +  2∠OAB = 1800                          [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25∘