Question

In the given figure, $Q$ is the centre of the circle and $PM$ and $PN$ are tangent segments to the circle.
If $m\angle MPN={70}^o$, 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

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

$\Rightarrow \angle MQN+{90}^o+{90}^o+{70}^o={360}^0$

$\therefore \angle MQN={360}^0-{250}^o={110}^o$