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 = 128°, then ∠OAB is equal to:

64°

As, PA and PB are tagents

\(\angle\)OAP = \({90}^\circ\)

\(\angle\)OBP = \({90}^\circ\)

As, OAPB is a quadrilateral

\(\angle\)OAP + \(\angle\)APB + \(\angle\)PBO + \(\angle\)BOA = 360

= \({90}^\circ\) + \({128}^\circ\) + \({90}^\circ\) + \(\angle\)BOA = 360

= \(\angle\)BOA = \({360}^\circ\) - \({308}^\circ\)

= \(\angle\)BOA = \({52}^\circ\)

As, OA = OB (Radii)

In \(\Delta \)OAB, \(\angle\)OBA + \(\angle\)BOA = 180

= 2 x \(\angle\)OAB + 52 = 180

= \(\angle\)OAB = (180 - 52)/2

= \(\angle\)OAB is \({64}^\circ\).