Question

In the given figure, $RQ\perp EF,AB\parallel CD$ and $EF$ is a transversal, find $x,y$ and $z$.

Answer 1

As $EF$ is a transversal, we can say the following:
$\angle APQ=\angle CQF$ (corresponding angles are equal) (1)
$\angle CQP+\angle CQF={180}^o$ (linear pair of angles) (2)
$x+y+z={180}^o$ (sum of angles of a triangle is ${180}^o$) (3)
Using (1):
${30}^o+y={65}^o⟹y={35}^o$
Using (2):
${65}^o+{25}^o+z={180}^o⟹z={90}^o$
Using (3):
${35}^o+{90}^o+x={180}^o⟹x={55}^o$
Final answer:
$x={55}^o,y={35}^o,z={90}^o$