Question

In the figure, PQ || RS, $\angle AEF={95}^0$, $\angle BHS={110}^0$ and $\angle ABC=x^0$. Then the value of x is

Answer 1

In the figure,

PQ || RS, $\angle AEF={95}^0$
$\angle BHS={110}^0,\angle ABC=x$
$\because$ PQ || RS,
$\therefore \angle AEF=\angle 1={95}^0$
(Corresponding angles)
But $\angle 1+\angle 2={180}^0$ (Linear pair)
$\Rightarrow \angle 2={180}^0-\angle 1={180}^0-{95}^0={85}^0$
In $\Delta$ AGH,
Ext. $\angle BHS=\angle +x$
$\Rightarrow {110}^0={85}^0+x$
$\Rightarrow x={110}^0-{85}^0={25}^0$