In figure, lines AB and CD intersect at O. If \(\angle{AOC}\) + \(\angle{BOE}\) = \(70^\circ\) and \(\angle{BOD}\) = \(40^\circ\), find \(\angle{BOE}\) and reflex \(\angle{COE}\).

Here, \(\angle{AOC}\) and \(\angle{BOD}\) are vertically opposite angles.

\(\therefore \angle{AOC}\) = \(\angle{BOD}\)
\(\because \) \(\angle{AOC}\) = \(40^\circ\)
\(\therefore \) \(\angle{BOD}\) = \(40^\circ\)] ....(i)
It is given that,
\(\angle{AOC} + \angle{BOE} = 70^\circ\)

Hence, from Eq. (i),
\(\because 40^\circ + \angle{BOE} = 70^\circ\)
\(\Rightarrow \angle{BOE} = 70^\circ - 40^\circ\)

\(\Rightarrow \) \(\angle{BOE}\) = \(30^\circ\)

Now, by Linear pair axiom,
\(\angle{AOC}\) + \(\angle{COE}\) + \(\angle{BOE}\) = \(180^\circ\)
By substituting the values, we get
\(40^\circ\) + \(\angle{COE}\) + \(30^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{COE}\) = \(180^\circ\) - \(40^\circ\) - \(30^\circ\)
\(\Rightarrow \) \(\angle{COE}\) = \(110^\circ\)

Now, so as to find the reflex angle,
\(\angle{COE}\) + reflex \(\angle{COE}\) = \(360^\circ\)
\(\Rightarrow \) \(110^\circ\) + reflex \(\angle{COE}\) = \(360^\circ\) .....(proved)
\(\Rightarrow \) reflex \(\angle{COE}\) = \(360^\circ\) - \(110^\circ\)
\(\Rightarrow \) reflex \(\angle{COE}\) = \(250^\circ\)