Question

In the given figure, $AB\parallel CD$ and $CD\parallel EF$. If $\angle BEF={64}^o$. Find the value of $x,y,z$

Answer 1

$\angle BEF={64}^o$

$AE$ is perpendicular to $EF$

$\therefore AEF={90}^o$

$=\angle AEB+\angle BEF$

${90}^o={64}^o+\angle AEB$

$z=\angle AEB={26}^o$

$\therefore ABE$ is a right angle trinalge

$\angle BAE+\angle AEB+\angle EBA={180}^o$

${90}^o+{26}^o+\angle EBA={180}^o$

$\angle EBA={180}^o-{116}^o={64}^o$