In the figure,lines AB and CD intersect at O. | vedclass

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(D) Since $AB$ is a straight line,the sum of angles on one side of the line at point $O$ is $180^o$.$	herefore \angle AOC + \angle COE + \angle BOE =

Vedclass · Accepted Answer

$30^o, 250^o$

Vedclass · Answer

(D) Since $AB$ is a straight line,the sum of angles on one side of the line at point $O$ is $180^o$.$	herefore \angle AOC + \angle COE + \angle BOE = 180^o$Rearranging the terms,we get:$(\angle AOC + \angle BOE) + \angle COE = 180^o$Given that $\angle AOC + \angle BOE = 70^o$,substituting this value:$70^o + \angle COE = 180^o$$\angle COE = 180^o - 70^o = 110^o$Reflex $\angle COE = 360^o - \angle COE = 360^o - 110^o = 250^o$Since lines $AB$ and $CD$ intersect at $O$,vertically opposite angles are equal:$\angle AOC = \angle BOD$Given $\angle BOD = 40^o$,therefore $\angle AOC = 40^o$.Using the given equation $\angle AOC + \angle BOE = 70^o$:$40^o + \angle BOE = 70^o$$\angle BOE = 70^o - 40^o = 30^o$Thus,$\angle BOE = 30^o$ and reflex $\angle COE = 250^o$.

In the figure,lines $AB$ and $CD$ intersect at $O$. If $\angle AOC + \angle BOE = 70^o$ and $\angle BOD = 40^o$,find $\angle BOE$ and reflex $\angle COE$.

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