Question

In the adjacent figure, $Q$ is the centre of the circle and $PM$ and $PN$ are tangent segments to the circle. If $m\angle MPN={60}^o$, find $\angle MQN$.

Answer 1

In MQNP, we can see that it is a quadrilateral hence sum of all the angles of the quadrilateral will be equal to ${360}^o$
As $\angle MPN+\angle PMQ+\angle PNQ+\angle MQN={360}^o$
$m\angle MPN={60}^0$ (Given)
$m\angle PMQ={90}^0$ (Radius is perpendicular to the tangent)
$m\angle PNQ={90}^0$ (Radius is perpendicular to the tangent)
$\therefore m\angle MQN={120}^0$ (Remaining angle)