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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