Question

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

• A. ${165}^0$
• B. ${60}^0$
• C. ${145}^0$
• D. ${120}^0$

Answer 1

The correct option is C ${145}^0$

Given that, AB ||CD ||EF and KL||HG.
Produce LK and GH as shown below:

$\because AB||CDandHK$ is transversal
$\therefore \angle 1=\angle DHK={25}^0$ (Alternate angles)
$\angle 3=\angle CHG={60}^0$ (Corresponding angles)
and $\angle 3=\angle 4$ (Corresponding angles)
= ${60}^0$
But $\angle 4+\angle 5={180}^0$ (Linear pair)
$\Rightarrow {60}^0+\angle 5={180}^0$
$\Rightarrow \angle 5={180}^0-{60}^0={120}^0$
$\therefore \angle HKL=\angle 1+\angle 5$
= ${25}^0+{120}^0={145}^0$