Two tangents PA and PB are drawn to a circle with centre O from an external point P. If ∠OAB = 30°, then ∠APB is:
Answer & explanation
Correct answer: option 4
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 \({60}^\circ\).