Question

In the figure, AB||CD||EF and GH||KL. Find $\angle HKL$.

Answer 1

In the figure, AB ||CD ||EF and KL||HG Produce LK and GH
$\because AB||CDandHK$ is transversal
$\therefore \angle 1={25}^0$ (Alternate angles)
$\angle 3={60}^0$ (Corresponding angles)
and $\angle 3=\angle 4$ (Corresponding angles)
= ${60}^0$
But $\angle 4+\angle 5={180}^0$ (Linear pair)
$\Rightarrow {60}^0+\angle 5={180}^0$
$\Rightarrow \angle 5={180}^0-{60}^0={120}^0$
$\therefore \angle HKL=\angle 1+\angle 5$
= ${25}^0+{120}^0={145}^0$