91Ó°ÊÓ

If two parallel lines are cut by transversal then same side interior angles are supplementary. Then,

$\begin{array}{c}3∠x+3∠y=180\\ 3\left(∠x+∠y\right)=180\\ ∠x+∠y=60°\end{array}$

The sum of the measure of all angles of a triangle is 180. Therefore,3

$\begin{array}{c}∠x+∠y+∠D=180\\ 60+∠D=180\\ ∠D=120°\end{array}$

The sum of the measure of all angles of a triangle is 180. Therefore,

$\begin{array}{c}2∠x+2∠y+∠B=180\\ 2\left(∠x+∠y\right)+∠B=180\\ 120+∠B=180\\ ∠B=60°\end{array}$

The sum of the measure of all angles of a triangle is 180. Therefore,

$\begin{array}{c}∠x+2∠y+∠A=180\\ ∠x+∠y+∠y+∠A=180\\ 60+∠y+∠A=180\\ ∠A=120°−∠y\end{array}$

The sum of the measure of all angles of a triangle is 180. Therefore,

$\begin{array}{c}2∠x+∠y+∠C=180\\ ∠x+∠x+∠y+∠C=180\\ 60+∠x+∠C=180\end{array}$

$∠C=120°−∠x$

The angles of ABCD is given by$∠A=120°−∠y,∠B=60°,∠C=120°−∠y$and$∠D=120°$