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 $\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{CD}$and $\stackrel{¯}{BD}$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.

$∠zâ‰\dots ∠ABD$

That is,$∠z=∠ABD=70°$

$∠AOD$and $∠BOC$are vertical angles, so by theorem 2-3 which states that the relationship between two angles is congruent when those angles are vertical angles,

$∠xâ‰\dots ∠AOD$

That is, $∠x=∠AOD=110°$

In the triangle $\mathrm{Δ}AOB$, $∠BOC$is the exterior angle corresponding to the two remote interior angles $∠BAO$and $∠ABO$, so by theorem 3-12 which states that the measurement related to an angle that is positioned on the exterior side of a triangle is the same as the measurement related to two angles that are remote interior angles when the remoter interior angles are added together,

$∠BAO+∠ABO=∠BOC$

$\begin{array}{c}y+70°=110°\\ y=110°−70°\\ y=40°\end{array}$