In the following figure,lines $AB$ and $CD$ intersect at $O.$ If $\angle AOC + \angle BOE = 100^{\circ}$ and $\angle BOD = 40^{\circ},$ find $\angle BOE$ and reflex $\angle COE.$

(N/A) Given that lines $AB$ and $CD$ intersect at $O.$
Since $AB$ is a straight line,the sum of angles on one side of the line is $180^{\circ}.$
Thus,$\angle AOC + \angle COE + \angle BOE = 180^{\circ}.$
We are given $\angle AOC + \angle BOE = 100^{\circ}.$
Substituting this into the equation: $100^{\circ} + \angle COE = 180^{\circ},$
which gives $\angle COE = 180^{\circ} - 100^{\circ} = 80^{\circ}.$
Since $AB$ and $CD$ are straight lines,vertically opposite angles are equal.
Therefore,$\angle AOC = \angle BOD = 40^{\circ}.$
Using the given equation $\angle AOC + \angle BOE = 100^{\circ},$
$40^{\circ} + \angle BOE = 100^{\circ},$
$\angle BOE = 100^{\circ} - 40^{\circ} = 60^{\circ}.$
Finally,the reflex $\angle COE = 360^{\circ} - \angle COE = 360^{\circ} - 80^{\circ} = 280^{\circ}.$