Question

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

Answer 1

In the figure, AB||CD ||EF and GH||KL and GH is produce to meet AB in L

$\because AB||CD$
$\therefore \angle 1={25}^0$ (Alternate angle)
and GH||KL
$\therefore \angle 4={60}^0$ (Corresponding angles)
$\angle 5=\angle 4={60}^0$ (Vertically opposite angle)
$\angle 5=\angle 2={180}^0$ (Co-interior angles)
$\Rightarrow {60}^0+\angle 2={180}^0$
$\therefore \angle 2={180}^0-{60}^0={120}^0$
Now $\angle HKL=\angle 1+\angle 2={25}^0+{120}^0$
=${145}^0$