Let O be the centre of a circle. PA and PB are tangents to the circle from a point P outside the circle and A and B are points on the circle, If angle APB = 50° , then angle OAB is equal to:

25°

∠OAP = ∠OBP  = 90º  ( Angle made by radius on circumference )

In quadrilateral OABP

∠O + ∠A + ∠B + ∠P = 360º

∠O + 90º+ 90º + 50º = 360º

∠O = 130º 

Now, in triangle OAB

∠OAB = ∠OBA ( Because OA = OB , angles made by equal sides are equal )

∠AOB + ∠OAB + ∠OBA = 180º

∠OAB + ∠OBA = 180º - 130º = 50º

2∠OAB = 50º

∠OAB = 25º ( ∠OAB = ∠OBA  )

∠OAB + ∠OBA = 180º