Question

In the given figure, O is the centre of the circle. If PA and PB are tangents from an external point P to the circle, then $\angle AQB$ is equal to

• A. ${100}^{\circ }$
• B. ${80}^{\circ }$
• C. ${70}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is D ${50}^{\circ }$

Tangents and radii are perpendicular at the point of contact.

In quadrilateral $OAPB$, we have:

$\angle APB={80}^o,\angle OAP=\angle OBP={90}^o$

By angle sum property of quadrilateral:

$\Rightarrow \angle AOB={100}^o$

Now, we know that angle subtended at the center is twice the angle subtended at any point on the arc, so we have:

$\angle AQB=\frac{1}{2}\angle AOB={50}^o$.

So option D is the right answer.