Question

In figure, $AB||CD||EF$ and $GH||KL$. The measure of $\angle HKL$ is

Answer 1

Given : $AB||CD||EF$ and $GH||KL$
Construction : Extend $GH$ to point $M$ such that $GM||KL$

$\therefore AB||>CD$ and $HK$ is transversal
$\therefore \angle DHK=\angle 1={25}^{\circ }\left(\text{Alternate interior angles}\right)$

$\because GH||KL$
$\therefore \angle CHG=\angle 4={60}^{\circ }\left(\text{Corresponding angles}\right)$
and $\angle 5=\angle 4={60}^{\circ }\left(\text{Vertically opposite angles}\right)$

Now, $\angle 5+\angle 2={180}^{\circ }\left(\text{Co-interior angles}\right)$
${60}^{\circ }+\angle 2={180}^{\circ }$
$\angle 2={180}^{\circ }-{60}^{\circ }$
$\angle 2={120}^{\circ }$

$\therefore \angle HKL=\angle 1+\angle 2={25}^{\circ }+{120}^{\circ }={145}^{\circ }$

Hence, option (c) is correct.