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Answer :

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 \(\angle{COE}\) = \(250^\circ\)

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