Lines and Angles — MATHS STD 7 — Question

 From the given figure, we can say that.
$\angle\text{BOC}=\angle\text{EOF} 40^\circ=\angle\text{EOF}$ [vertycally opposite angles]
$\Rightarrow40^\circ=\angle\text{EOF}$
Since, sum of all the angles on a straight line is $180^\circ $
$\therefore\angle\text{BOC} +\angle\text{FOE}+\angle\text{EOD}=180^\circ$
$\Rightarrow 90^\circ+40^\circ+5\text{a}=180^\circ$
$\Rightarrow130^\circ+5\text{a}=180^\circ\Rightarrow5\text{a}=180^\circ-130^\circ$
$\Rightarrow5\text{a}=50^\circ$
$\Rightarrow\text{a}=\frac{50^\circ}{5}=10^\circ$