Question

$\text{From the given figure, find}\angle AOB\text{(in degrees )}.$ [2 Marks]

Answer 1

We know that, tangents drawn at any point on the circle is perpendicular to the radius through the point of contact.

$\therefore OA\perp PA\text{and}OB\perp PB\Rightarrow \angle OAP=\angle OBP={90}^{\circ }$
[1 Mark]
$\text{Also, the sum of all four angles of}\text{quadrilateral is}{360}^{\circ }.$

$\therefore \angle OAP+\angle APB+\angle OBP+\angle AOB={360}^{\circ }\Rightarrow {90}^{\circ }+{40}^{\circ }+{90}^{\circ }+\angle AOB={360}^{\circ }\Rightarrow \angle AOB={360}^{\circ }-{90}^{\circ }-{40}^{\circ }-{90}^{\circ }\therefore \angle AOB={140}^{\circ }$
[1 Mark]