In the given figure, if $A B \| C D, \angle A P Q=50^{\circ}$ and $\angle P R D=127^{\circ}$, find $x$ and $y$.

Solution:

$\angle \mathrm{APR}=\angle \mathrm{PRD}$ (Alternate interior angles)

$50^{\circ}+y=127^{\circ}$

$y=127^{\circ}-50^{\circ}$

$y=77^{\circ}$

Also, $\angle \mathrm{APQ}=\angle \mathrm{PQR}$ (Alternate interior angles)

$50^{\circ}=x$

$\therefore x=50^{\circ}$ and $y=77^{\circ}$