Question

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

( 2 marks)

Answer 1

By Theorem- Tangents drawn at any point on the circle is perpendicular to the radius through the point of contact.
$\therefore OA\perp PA$ and $OB\perp PB$
$\Rightarrow \angle OAP=\angle OBP={90}^{\circ }$ (1 mark)

By Theorem- Sum of all four angles of quadrilateral = ${360}^{\circ }$
$\therefore \angle OAP+\angle APB+\angle OBP+\angle AOB={360}^{\circ }$
$\Rightarrow {90}^{\circ }+{40}^{\circ }+{90}^{\circ }+\angle AOB={360}^{\circ }$
$\Rightarrow \angle AOB={360}^{\circ }-{90}^{\circ }-{40}^{\circ }-{90}^{\circ }$
$\therefore \angle AOB={140}^{\circ }$ (1 mark)