Two tangents PA and PB are drawn to a circle with centre O from an external point P. If ∠OAB = 30°, then ∠APB is:

60°

Given,

\(\angle\)OAB = \(\angle\)OBA = \({30}^\circ\)   [OA = OB  (radius)]

As we know,

\(\angle\)OAP = \(\angle\)OBP = \({90}^\circ\)   [tangents]

= \(\angle\)OAB + \(\angle\)BAP = \({90}^\circ\)

= 30 + \(\angle\)BAP = \({90}^\circ\)

= \(\angle\)BAP = \({60}^\circ\)

Similarily

\(\angle\)OBA + \(\angle\)ABP = \({90}^\circ\)

= \({30}^\circ\) + \(\angle\)ABP = \({90}^\circ\)

= \(\angle\)ABP = \({60}^\circ\)

In \(\Delta \)ABP

\(\angle\)BAP + \(\angle\)ABP + \(\angle\)APB = \({180}^\circ\)

= \({60}^\circ\) + \({60}^\circ\) + \(\angle\)APB = \({180}^\circ\)

= \(\angle\)APB = \({180}^\circ\) - \({120}^\circ\) = \({60}^\circ\).

Therefore, \(\angle\)APB is \({60}^\circ\).