In Fig. $10.315$, $AB||CD||EF$ and $GH||KL$. Find $\angle HKL$."

Let us extend the line $GH$ to $N$ such that it intersects the line $AB$ at $M$

$AB||CD$ and $GM$ is transversal 

 

$\Rightarrow The \angle AMH=\angle CHG=60^{\circ}$  [Corresponding Angles]

$\angle NMB=\angle AMH=60^{\circ}$          [Vertically opposite angles]

$AB∣∣CD$ and $HK$ is transversal 

$\Rightarrow  The \angle AKH=\angle KHD=25^{\circ}$     [Alternate interior Angles]

$NM||KL$ and $KM$ is transversal 

 

$\Rightarrow  The \angle NMK=\angle LKB=60^{\circ}$     [Corresponding Angles]

$\Rightarrow \angle LKM=180^{\circ}-\angle KLB$             [Linear pair]

$\Rightarrow \angle LKM=180^{\circ}-60^{\circ}=120^{\circ}$

 

$\Rightarrow \angle HKL=\angle HKA+\angle AKL$

$\Rightarrow \angle HKL=25^{\circ}+120^{\circ}=145^{\circ}$