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:

$71^\circ$

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 \({71}^\circ\).