Use the figure shown to find $m\angle WXY$ and $m\angle XYZ$. Explain your reasoning.

The figure is shown below.

Angle sum property of quadrilateral:

$\angle 1+\angle 2+\angle 3+\angle 4=360°$

Alternate angle property.

Let’s name the points of the triangles as A and B.

In$\Delta WAX$:

To find angle$\angle WXA$, substituting the value of angles,

$\begin{array}{c}45+\angle WXA+90=180\\ \angle WXA+135=180\end{array}$

Shifting left-hand side value to right-hand side:

$\begin{array}{c}\angle WXA=180-135\\ \angle WXA=45\end{array}$

In $\Delta YBZ$:

To find angle$\angle BYZ$, substituting the value of angles,

$\begin{array}{c}60+\angle BYZ+90=180\\ \angle BYZ+150=180\end{array}$

Shifting left-hand side value to right-hand side:

$\begin{array}{l}\angle BYZ=180-150\\ \angle BYZ=30\end{array}$

Considering the alternate angle property, which states that when the lines are parallel, that the alternate interior angles are equal. Hence,

$\angle XAW=\angle YXA$,

$\angle YXA=90$, and

$\angle ZBY=\angle XYB$

$\angle XYB=90$

For$\angle WXY$,

$\begin{array}{c}\angle WXY=\angle YXA+\angle WXA\\ \angle WXY=45+90\\ \angle WXY=135\end{array}$

For$\angle XYZ$,

$\begin{array}{c}\angle XYZ=\angle XYB+\angle ZYB\\ \angle XYZ=30+90\\ \angle WXY=120\end{array}$

Hence,

$\begin{array}{c}\angle WXY=135°\\ \angle XYZ=120°\end{array}$