In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O.

We know that if two lines intersect, then the vertically-opposite angles are equal.

$\therefore \angle D O F=\angle C O E=5 x^{\circ}$

$\angle A O D=\angle B O C=2 x^{\circ}$ and

$\angle A O E=\angle B O F=3 x^{\circ}$

Since, AOB is a straight line, we have:

$\angle A O E+\angle C O E+\angle B O C=180^{\circ}$

$\Rightarrow 3 x+5 x+2 x=180^{\circ}$

$\Rightarrow 10 x=180^{\circ}$

$\Rightarrow x=18^{\circ}$

Therefore,

$\angle A O D=2 \times 18^{\circ}=36^{\circ}$

$\angle C O E=5 \times 18^{\circ}=90^{\circ}$

$\angle A O E=3 \times 18^{\circ}=54^{\circ}$