Question

In the given figure , Q is the centre of the circle and PM and PN are the tangent segments to the circle.
If $\angle$MPN = 40$°$, find $\angle$MQN

Answer 1

$$\begin{gathered} \angle \mathrm{QMP}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \angle \mathrm{QNP}=90°(\mathrm{Tangent}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}) \\ \mathrm{In}\mathrm{quadrialteral}\mathrm{PMQN},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{QMP}+\angle \mathrm{QNP}+\angle \mathrm{MPN}+\angle \mathrm{MQN}=360°(\therefore \mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle \mathrm{MQN}=360°-\angle \mathrm{QMP}-\angle \mathrm{QNP}-\angle \mathrm{MPN} \\ \Rightarrow \angle \mathrm{MQN}=360°-90°-90-40°=140° \end{gathered}$$