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:
Answer & explanation
Correct answer: option 4
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 \({64}^\circ\).