Question

In figure, if PQR is tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and $\angle BQR={70}^{\circ }$, then $\angle AQB$ is equal to:

• A. ${20}^{\circ }$
• B. ${40}^{\circ }$
• C. ${35}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${40}^{\circ }$

$\angle ABQ=\angle BQR={70}^{\circ }$ [$\therefore AB||PR$]
$\angle OQB={90}^{\circ }-{70}^{\circ }={20}^{\circ }$
Now, $\angle OQB=\angle OBQ={20}^{\circ }$
$\therefore \angle OBA=\angle OAB={50}^{\circ }$
So, $\angle AOB={180}^{\circ }-({50}^{\circ }+{50}^{\circ })={80}^{\circ }$
$\angle AQB=\frac{1}{2}\angle AOB$
$=\frac{1}{2}\left({80}^{\circ }\right)={40}^{\circ }$