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:
Answer & explanation
Correct answer: option 1
∠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