In Fig. 6.32, if $ \mathrm{AB} \| \mathrm{CD}, \angle \mathrm{APQ}=50^{\circ} $ and $ \angle \mathrm{PRD}=127^{\circ} $, find $ x $ and $ y $.
"

Given:

$AB \parallel CD$, $\angle APQ=150^0$ and $\angle PRD=127^o$.

To do:

We have to find $x$ and $y$.

Solution:

We know that,

If the lines intersected by the transversal are parallel, alternate interior angles are equal.

This implies,

$\angle APQ=\angle PQR$

By substituting the values we get,

$\angle APQ=\angle PRD$

This implies,

$x=50^o$

In the similar way, we get,

$\angle APR=\angle PRD$

By substituting the value of $\angle PRD$

We get,

$\angle APR=127^o$

We know that,

$\angle APR=\angle APQ+\angle QPR$

Now, by substituting the values of $\angle QPR=y$ and $\angle APR=127^o$

We get,

$127^o=50^o+y$

This implies,

$127^o-50^o=y$

$77^o=y$

Therefore,

$y=77^o$

Hence, the values of $x$ and $y$ are $50^o$ and $77^o$ respectively.