Question

In the given figure, PA and PB are two tangents drawn from an external point P. If $\angle APB={40}^{\circ }$, then $\angle PAB=$

• A. ${70}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$

Answer 1

The correct option is A ${70}^{\circ }$

By Theorem- If two tangents PA and PB are drawn to a circle with centre 'O' from an external point P, then $\angle APB=2\angle OAB=2\angle OBA$
$\Rightarrow \angle OAB=\frac{APB}{2}$

$\therefore \angle OAB=\frac{{40}^{\circ }}{2}={20}^{\circ }$

By Theorem- Tangent drawn at any point to the circle is perpendicular to the radius through the point of contact
$\therefore OA\perp PA$
$\Rightarrow \angle OAP={90}^{\circ }$

$\because \angle OAP=\angle OAB+\angle PAB$
$\therefore {90}^{\circ }={20}^{\circ }+\angle PAB$
$\Rightarrow \angle PAB={70}^{\circ }$