Mathematics

Three lines $AB,\ CD$ and $EF$ intersect each other at $O$. If $\angle AOE=30^o$ and $\angle DOB=40^o$ Fig. find $\angle COF$.

SOLUTION
As $AB$ is a straight line.
$\therefore \angle AOE + \angle EOD +\angle DOB=180^o$
$\Rightarrow \angle EOD=180^o-30^o-40^o$
$\Rightarrow \angle EOD=110^o \ .....(1)$
Now, $CD$ and $EF$ intersect each other at point $O$.
$\therefore \angle COF =\angle EOD =110^o$ [vertically opposite angles]
Therefore, $\angle COF=110^o$

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Subjective Medium Published on 09th 09, 2020
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