In the given figure, AB || CD. Find the value of x.

$A B \| C D$ and $\mathrm{PQ}$ is the transversal.

Then,

$\angle P E F=\angle E G H \quad[$ Corresponding Angles $]$

$\Rightarrow \angle E G H=85^{\circ}$

And,

$\angle E G H+\angle Q G H=180^{\circ}$

$\Rightarrow 85^{\circ}+\angle Q G H=180^{\circ}$

$\Rightarrow \angle Q G H=95^{\circ}$

Also,

$\angle C H Q+\angle G H Q=180^{\circ} \quad$ [Since CD is a straight line]

$\Rightarrow 115^{\circ}+\angle G H Q=180^{\circ}$

$\Rightarrow \angle G H Q=65^{\circ}$

We know that the sum of angles of a triangle is $180^{\circ}$.

$\Rightarrow \angle Q G H+\angle G H Q+\angle G Q H=180^{\circ}$

$\Rightarrow 95^{\circ}+65^{\circ}+x=180^{\circ}$

$\Rightarrow x=20^{\circ}$

$\therefore x=20^{\circ}$