91Ó°ÊÓ

According to the provided information, the opposite sides of the provided figure are parallel. That is$\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{CD}$ and $\stackrel{¯}{AD}âˆ\yen \stackrel{¯}{BC}$.

Now, lines$\stackrel{¯}{AD}âˆ\yen \stackrel{¯}{BC}$ and $\stackrel{¯}{AB}$is transversal, so by Theorem 3-2 which states that if two lines are having a separate line that is travelling through them then the angles are congruent which are positioned on the alternate interior sides of the lines.

Write the alternate interior angles.

$∠zâ‰\dots ∠DBC$

That is, $∠z=∠DBC=85°$

Also since ABCD is a parallelogram, so by theorem 5-2 states that the relationship between two angles that are related to a parallelogram and positioned on the opposite sides of each other is congruent.

Write the opposite angles of provided figure.

$∠xâ‰\dots ∠DAB$

That is,$∠x=∠DAB=56°$

Now consider the triangle $\mathrm{Δ}DAB$, so by theorem 3-11 which states that if all the angles that are positioned in a triangle are added together then they will result in $180°$.

$∠ADB+∠DBA+∠BAD=180°$

$\begin{array}{c}85°+y+56°=180°\\ y+141°=180°\\ y=180°−141°\\ y=39°\end{array}$