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 $\angle APB = 142^\circ$, then $\angle OAB$ is equal to:
Answer & explanation
Correct answer: option 3
As, PA and PB are tangents
\(\angle\)OAP = 90
\(\angle\)OBP = 90
As, OAPB is a quadrilateral
\(\angle\)OAP + \(\angle\)APB + \(\angle\)PBO + \(\angle\)BOA = 360
= 90 + 142 + 90 + \(\angle\)BOA = 360
= \(\angle\)BOA = 360 - 322
= \(\angle\)BOA = 38
As, OA = OB (Radii)
In \(\Delta \)OAB, \(\angle\)OBA (In a triangle angles opposite to equal sides are equal)
\(\angle\)OAB + \(\angle\)OBA + \(\angle\)BOA = 180
= 2 x \(\angle\)OAB + 38 = 180
= \(\angle\)OAB = \(\frac{180\; -\; 38}{2}\)
= \(\angle\)OAB is \({71}^\circ\).