PA and PB are two tangents from a point P outside the circle with centre O at the points A and B on it. If ∠APB = 130°, then ∠OAB is equal to:
Answer & explanation
Correct answer: option 4
\(\angle\)OAB = \(\angle\)OBA = \(\frac{1}{2}\) x \(\angle\)APB
= \(\angle\)OAB = \(\frac{1}{2}\) x 130 = \({65}^\circ\)
Therefore, \(\angle\)OAB is \({65}^\circ\).