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# 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 $$\angle{COE}$$ = $$250^\circ$$