Point $\mathrm{P}$ lies outside a circle with centre $\mathrm{O}$. Tangents $\mathrm{PA}$ and $\mathrm{PB}$ are drawn to meet the circle at $\mathrm{A}$ and $\mathrm{B}$ respectively. If $\angle \mathrm{APB}=80^{\circ}$, then $\angle \mathrm{OAB}$ is equal to:
Answer & explanation
Correct answer: option 2
In the given figure,
BOAP is a quadrilateral, \(\angle\)OAP and \(\angle\)OBP = \({90}^\circ\)
So, \(\angle\)AOB + \(\angle\)APB + \({90}^\circ\) + \({90}^\circ\) = \({360}^\circ\)
= \(\angle\)AOB = \({100}^\circ\)
In an \(\Delta \)OAB,
OA = OB = radius
= \(\angle\)OAB =\(\angle\)OBA
= \(\angle\)OAB = \(\frac{180\;-\;100}{2}\) = \(\frac{80}{2}\) = \({40}^\circ\)
Therefore, \(\angle\)OAB = \({40}^\circ\).