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

Consider the figure.

Here, AB || CD and a transversal is cutting them. 

Now, 

 $\angle \mathrm{FEP}=\angle \mathrm{HGE}$                  [Corresponding angles are equal]

$\angle \mathrm{HGE}={85}^0$  

The angles $\angle \mathrm{HGE}, \angle \mathrm{HGQ}$ are linear pairs, so 
$\angle \mathrm{HGE}+\angle \mathrm{HGQ}=180°$ 
${85}^0+\angle \mathrm{HGQ}=180°$ 
$\angle \mathrm{HGQ}=95°$ 
In $\mathrm{\Delta GQH}$, $\angle \mathrm{CHQ}$ is the exterior angles. 
$\Rightarrow \angle \mathrm{CHQ}=\angle \mathrm{HGQ}+\angle \mathrm{GQH}$ 
$\Rightarrow 115°=95°+\mathrm{x}$ 
$\Rightarrow \mathrm{x}=20°$  
The value of x is $20°$.