# 1. In Fig. 6.13, lines AB and CD intersect at O. If $\angle$ AOC + $\angle$ BOE = 70° and $\angle$ BOD = 40°, find $\angle$BOE and reflex $\angle$COE.

M manish painkra

Given that,
AB is a straight line. Lines AB and CD intersect at O. $\angle AOC + \angle BOE = 70^0$ and $\angle$ BOD = $40^0$
Since AB is a straight line
$\therefore$ $\angle$AOC + $\angle$COE + $\angle$EOB = $180^0$
$\Rightarrow \angle COE = 180^0-70^0=110^0$ [since $\angle AOC + \angle BOE = 70^0$]

So, reflex $\angle$COE = $360^0-110^0 = 250^0$
It is given that AB and CD intersect at O
Therefore, $\angle$AOC  = $\angle$BOD  [vertically opposite angle]
$\Rightarrow \angle COA = 40^0$ [ GIven $\angle$ BOD = $40^0$]
Also, $\angle AOC + \angle BOE = 70^0$
So,  $\angle$BOE = $30^0$

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