Question

In the fig. 1, PA and PB are tangents to the circle with centre O such that $\angle APB={50}^{\circ }$. Write the measure of $\angle OAB$.

Answer 1

PA and PB are tangents drawn from an external point P to the circle.
$\therefore$ PA = PB (Length of tangents drawn from an external point to the circle are equal)
In $\Delta PAB$,
PA = PB
$\Rightarrow \angle PBA=\angle PAB$ ....(1) (Angles opposite to equal sides are equal).
Now,
$\angle APB+\angle PBA+\angle PAB={180}^{\circ }$
$\Rightarrow {50}^{\circ }+\angle PAB+\angle PAB={180}^{\circ }\left[Using\right(1\left)\right]$
$\Rightarrow 2\angle PAB={130}^{\circ }$
$\Rightarrow 2\angle PAB=\frac{{130}^{\circ }}{2}={65}^{\circ }$
We know that radius is perpendicular to the tangent at the point of contact
$\therefore \angle OAP={90}^{\circ }(OA\perp PA)$
$\Rightarrow \angle PAB+\angle OAB={90}^{\circ }$
$\Rightarrow {65}^{\circ }+\angle OAB={90}^{\circ }$
$\Rightarrow \angle OAB={90}^{\circ }-{65}^{\circ }={25}^{\circ }$
Hence the measure of $\angle OAB$ is ${25}^{\circ }$