PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 142°, then ∠OAB is equal to:

Answer: Option C

Solution

Answer: Option C

Solution:

$$\angle \text{OAB}=\frac{{180}^{\circ }-{38}^{\circ }}{2}=71^{\circ }$$