Question

In the following figure, $Q$ is the centre of circle and $PM$ and $PN$ are tangent segments to the circle. If $\angle MPN={40}^{\circ }$, find $\angle MQN$

Answer 1

Since angle between tangent to a circle and the radius passing through the point of contact is ${90}^o$

so $\angle PNQ=\angle PMQ={90}^o$,

Now since $PNQM$ is forming a quadrilateral, so sum of its all inernal angle will be ${360}^o$

Hence

$\angle MQN+\angle PNQ+\angle PMQ+\angle MPN={360}^0$

$\Rightarrow \angle MQN+{90}^o+{90}^o+{40}^o={360}^0$

$\therefore \angle MQN={360}^0-{220}^o={140}^o$