Question
Three lines $AB, CD$ and $EF$ intersect each other at $O.$ If $\angle\text{AOE} = 30^\circ$ and $\angle\text{DOB} = 40^\circ$ Fig. find $\angle\text{COF}$.
✓
Answer
From the given figure, we have.
$\angle\text{AOE}+\angle\text{EOD}+\angle\text{DOB}=180^\circ[\because$ sum of all the angles on a straight line is $180^\circ ]$
$\Rightarrow30^\circ+\angle\text{EOD}+40^\circ=180^\circ$
$\Rightarrow\angle\text{EOD}=180^\circ-70^\circ$
$\Rightarrow\angle\text{EOD}=110^\circ$
Again, $\angle\text{EOD}=\angle\text{COF} [$Vertycally opposite angles$]$
$\Rightarrow\angle\text{COF}=110^\circ$
$\angle\text{AOE}+\angle\text{EOD}+\angle\text{DOB}=180^\circ[\because$ sum of all the angles on a straight line is $180^\circ ]$
$\Rightarrow30^\circ+\angle\text{EOD}+40^\circ=180^\circ$
$\Rightarrow\angle\text{EOD}=180^\circ-70^\circ$
$\Rightarrow\angle\text{EOD}=110^\circ$
Again, $\angle\text{EOD}=\angle\text{COF} [$Vertycally opposite angles$]$
$\Rightarrow\angle\text{COF}=110^\circ$
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