Question

In figure, if PA and PB are tangents to the circle with centre O such that $\angle APB={50}^{\circ }$, then $\angle OAB$ is equal to:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given that PA and PB are tangent lines.
PA= PB [ Since , the length of tangents drawn from an external point to a circle is equal ]
$\Rightarrow$ $\angle PBA=\angle PAB=\theta$
In $\Delta PAB,\angle P+\angle A+\angle B={180}^{\circ }$ [Angle sum property of triangle]
$\Rightarrow$ ${50}^{\circ }+\theta +\theta ={180}^{\circ }$
$\Rightarrow$ $2\theta ={180}^{\circ }-{50}^{\circ }={130}^{\circ }$
$\Rightarrow$ $\theta ={65}^{\circ }$
Also, OA $\perp$ PA
[Since, tangents at any point of a circle is perpendicular to the radius through the point of contact]
$\therefore$ $\angle PAO={90}^{\circ }$
$\Rightarrow$ $\angle PAB+\angle BAO={90}^{\circ }$
$\Rightarrow$ ${65}^{\circ }+\angle BAO={90}^{\circ }$
$\Rightarrow$ $\angle BAO={90}^{\circ }-{65}^{\circ }={25}^{\circ }$