Question

$AB$ is parallel to $CD.EF$ intersects them at $M$ and $N$. The bisectors of $\angle M$ and $\angle N$ meet at $Q$. If $\angle AME={80}^{\circ }$, then $\angle MQN$ is equal to

• A. ${60}^{\circ }$
• B. ${70}^{\circ }$
• C. ${80}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

$\angle BMN=\angle AME={80}^{\circ }(vert.opp.\angle sareequal)$
$\angle BMN+\angle MND={180}^{\circ }(AB\parallel CD,co-int.\angle saresupp)$
$\Rightarrow \angle MND={180}^{\circ }-\angle BMN$
$={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
$MQ$ bisects $\angle BMN\Rightarrow \angle NMQ={40}^{\circ }$
$NQ$ bisects $\angle MND\Rightarrow \angle QNM={50}^{\circ }$
$\therefore$ In $\Delta MQN$,
$\angle MQN+\angle NMQ+\angle QNM={180}^{\circ }$
$\Rightarrow \angle MQN+{40}^{\circ }+{50}^{\circ }={180}^{\circ }$
$\Rightarrow \angle MQN={180}^{\circ }-{90}^{\circ }={90}^{\circ }$.