In $\Delta ABC$, $BD\perp AC$ at $D$. $E$ is a point on $BC$ such that $\angle BEA=x^{\circ }$. If $\angle EAC=36^{\circ }$ and $\angle EBD=40^{\circ }$, then the value of $x$ is:

In $\Delta BDC$, we are given $BD\perp AC$, so $\angle BDC=90^{\circ }$.In right-angled $\Delta BDC$:
$$\angle CBD+\angle BCD+\angle BDC=180^{\circ }$$
Since $\angle EBD=40^{\circ }$ and $E$ lies on $BC$, $\angle CBD=40^{\circ }$.
$$40^{\circ }+\angle C+90^{\circ }=180^{\circ }⟹\angle C=50^{\circ }$$
Now, in $\Delta AEC$, the exterior angle $\angle BEA$ is equal to the sum of the two interior opposite angles:
$$\angle BEA=\angle EAC+\angle ECA$$
Given $\angle EAC=36^{\circ }$ and $\angle ECA=\angle C=50^{\circ }$:
$$x^{\circ }=36^{\circ }+50^{\circ }=86^{\circ }$$
Therefore, $x=86$.