Question

In the figure, AB is parallel to CD, if $\angle 1+\angle 3={240}^{\circ }$, find the value of $\angle 6$.

Answer 1

$\angle 1+\angle 3={240}^{\circ }$ - Given
Also , $\angle 1=\angle 3$ because they are vertically opposite angles.
Therefore, $2\times \angle 3={240}^{\circ }$
$\angle 3={120}^{\circ }$

Since $\angle 3$ and $\angle 6$ are co-interior angles,
$\angle 3+\angle 6={180}^{\circ }$

$\angle 6={180}^{\circ }-{120}^{\circ }={60}^{\circ }$