Question

If AB || CD and XY is a transversal which meets AB at M & CD at N and also $\angle AMX:\angle BMX=3:2$, then $\angle CNY=$

• A. ${36}^{\circ }$
• B. ${72}^{\circ }$
• C. ${108}^{\circ }$

Answer 1

The correct option is B ${72}^{\circ }$

Given $\angle 1:\angle 2=3:2$
$\therefore \angle 1=\frac{3}{5}\times 180={108}^{\circ }$
$\angle 5=\angle 1$ (corresponding angles)
$\angle 5+\angle 7={180}^{\circ }$ (linear pair)

Therefore, $\angle 7=\angle CNY={180}^{\circ }-{108}^{\circ }={72}^{\circ }$