Question

In figure, $AB||CD||EFandGH||KL$. Find $\angle HKL.$

Answer 1

Given: $AB||CD||EFandGH||KL$
Construction: Produce $GH$ to point $M$ such that $GM||KL$

$\because AB||CDandHK$ is transversal

$\therefore \angle DHK=\angle 1={25}^0$ $\left(\text{alternate interior angles}\right)$

$\therefore \angle CHG=\angle 4={60}^0$ $\left(\text{corresponding angles}\right)$

$\angle 5=\angle 4={60}^0$ $\left(\text{vertically opposite angles}\right)$

$\because GM||KL$

$\angle 5+\angle 2={180}^0$ $\left(\text{co-interior angles}\right)$

${60}^0+\angle 2={180}^0$

$\angle 2={180}^0-{60}^0$

$\angle 2={120}^0$

$\therefore \angle HKL=\angle 1+\angle 2={25}^0+{120}^0={145}^0$

Hence, $\angle HKL={145}^0$.