Question

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

• A. ${140}^{\circ }$
• B. ${110}^{\circ }$
• C. ${150}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is A ${140}^{\circ }$

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

$\therefore OA\perp PA\text{and}OB\perp PB\Rightarrow \angle OAP=\angle OBP={90}^{\circ }$

$\text{Also, Sum of all four angles of}\text{quadrilateral is}{360}^{\circ }.$

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