Question

In the given figure, PA and PB are two tangents to the circle with center O. If $\angle APB={40}^{\circ }$, find $\angle AQB$ and $\angle AMB$. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

We know that $\angle OAP=\angle OBP={90}^{\circ }$

$\angle AOB={360}^{\circ }-\angle OAP-\angle OBP-\angle APB$

$\angle AOB={360}^{\circ }-{90}^{\circ }-{90}^{\circ }-{40}^{\circ }$

$\angle AOB={140}^{\circ }$

We know that $\angle AQB=\frac{1}{2}\angle AOB$

(Angle bisected by a chord on the center of the circle is twice the angle bisected on the circle)

$\angle AQB=\frac{1}{2}{140}^{\circ }$

$\angle AQB={70}^{\circ }$

$\angle AMB={180}^{\circ }-\angle AQB$ (Cyclic quadrilateral)

$\angle AMB={180}^{\circ }–{70}^{\circ }$

$\angle AMB={110}^{\circ }$

Therefore $\angle AQB={70}^{\circ }$ and $\angle AMB={110}^{\circ }$