Correct Answer: $10^{\circ}$

Solution :
Let AD be the angle bisector and AE be the perpendicular.
⇒ $\angle$BAC + $\angle$ABC + $\angle$ACB = $180^{\circ}$
⇒ $\angle$BAC + $60^{\circ}$ + $40^{\circ}$ = $180^{\circ}$
⇒ $\angle$BAC = $180^{\circ}$ – $100^{\circ}$ = $80^{\circ}$
⇒ $\angle$BAD = $\frac{80^{\circ}}{2}$ = $40^{\circ}$
In $\triangle$AEB,
⇒ $\angle$BAE + $\angle$ABE + $\angle$AEB = $180^{\circ}$
⇒ $\angle$BAE + $60^{\circ}$ + $90^{\circ}$ = $180^{\circ}$
⇒ $\angle$BAE = $180^{\circ}$ – $150^{\circ}$ = $30^{\circ}$
⇒ $\angle$BAE = $30^{\circ}$
$\therefore$ $\angle$EAD = $\angle$BAD – $\angle$BAE = $40^{\circ}$ – $30^{\circ}$ = $10^{\circ}$
Hence, the correct answer is $10^{\circ}$.