Question

If $AB$ is parallel to $CD$, then $\angle MON=$

• A. ${40}^{\circ }$
• B. ${40}^{\circ }$
• C. ${60}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$


From the image, $AB$ is parallel to $CD$
$\Rightarrow \angle NQD=\angle ONB={120}^{\circ }$ (Corresponding angles)
Also, $\angle AMO+\angle OMN={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle OMN={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
Also, $\angle ONM+\angle ONB={180}^{\circ }$ (Linear pair)

$\Rightarrow \angle ONM={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
Now, $OMN$ forms a triangle.
$\Rightarrow \angle OMN+\angle ONM+\angle MON={180}^{\circ }$
Angle sum property of a triangle)
$\Rightarrow {60}^{\circ }+{60}^{\circ }+\angle MON={180}^{\circ }$
$\Rightarrow \angle MON={180}^{\circ }-{120}^{\circ }={60}^{\circ }$