Question

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

• A. ${25}^{\circ }$
• B. ${30}^{\circ }$
• C. ${40}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is D ${130}^{\circ }$

In $\square PAOB$,

$\angle APB+\angle PBO+\angle AOB+\angle PAO={360}^{\circ }$ ....Angle sum property of quadrilateral

$\Rightarrow \angle AOB={360}^{\circ }-({50}^{\circ }+{90}^{\circ }+{90}^{\circ })={130}^{\circ }$

So, option D is correct.