Point A and B are on a circle with centre O. PA and PB are tangents to the circle from an external point P. If PA and PB are inclined to each other at 42°, then find the measure of ∠OAB.

21°

Consider the quadrilateral AOBP,

⇒ \(\angle\)A + \(\angle\)O + \(\angle\)B + \(\angle\)P = \({360}^\circ\)

⇒ \({90}^\circ\) + \(\angle\)O + \({90}^\circ\) + \({42}^\circ\) = \({360}^\circ\)

⇒ \(\angle\)O = \({360}^\circ\) -  \({90}^\circ\) - \({90}^\circ\) - \({42}^\circ\)

⇒ \(\angle\)O = \({138}^\circ\)

⇒ \(\Delta \)OAB is an isosceles triangle since OA and OB are the radii of the circle.

So, \(\angle\)OAB = \(\angle\)OBA = \({R}^\circ\)  (let)

⇒  2\({R}^\circ\) + \({138}^\circ\) = \({180}^\circ\)

⇒ 2\({R}^\circ\) = \({42}^\circ\)

⇒ R = \({21}^\circ\)

⇒ \(\angle\)OAB = \({21}^\circ\)

Therefore, \(\angle\)OAB is \({21}^\circ\).