, 2026

In equilateral $\$\triangle$
PXY$, the measure of$$\angle$
PXY$is$$60\degree$$and$PX=XY$.

In square $WXYZ$, the measure of
$\angle WXY$ is $90\degree$ and $WX=XY$.

Therefore, $\$\angle WXP=\angle WXY-\angle PXY=90\degree-60\degree=30\degree$$and$WX=PX$.

Thus $\triangle WXP$ is isosceles,
and so $\$\angle PWX=\angle WPX=\dfrac{180\degree-30\degree}{2}=75\degree\$.$

The measure of $\angle PWZ$ is $\$90\degree-\angle PWX=90\degree-75\degree=15\degree\$.$

In a similar way, we can show that $\$\angle PZW=15\degree$$, and so$$\angle WPZ=180\degree-\angle PWZ-\angle$ PZW$or$$\angle WPZ=180\degree-15\degree-15\degree=150\degree\$.$