Question
In the adjoining figure, three coplanar lines $AB, CD$ and $EF$ intersect at a point $O$. Find the value of $x$. Also, find $\angle\text{AOD},\angle\text{COE}$ and $\angle\text{AOE}.$
✓
Answer
We know that if two lines intersect, then the vertically-opposite angles are equal.
$\therefore\angle\text{DOF}=\angle\text{COE}=5\text{x}^\circ$ $\angle\text{AOD}=\angle\text{BOC}=2\text{x}^\circ\text{ and}$ $\angle\text{AOE}=\angle\text{BOF}=3\text{x}^\circ$
Since, AOB is a straight line, we have: $\angle\text{AOE}+\angle\text{COE}+\angle\text{BOC}=180^\circ$
$\Rightarrow 3x + 5x + 2x = 180^\circ$
$\Rightarrow 10x = 180^\circ$
$\Rightarrow x = 18^\circ$
Therefore, $\angle\text{AOD}=2\times18^\circ=36^\circ$
$\angle\text{COE}=5\times18^\circ=90^\circ$
$\angle\text{AOE}=3\times18^\circ=54^\circ$
$\therefore\angle\text{DOF}=\angle\text{COE}=5\text{x}^\circ$ $\angle\text{AOD}=\angle\text{BOC}=2\text{x}^\circ\text{ and}$ $\angle\text{AOE}=\angle\text{BOF}=3\text{x}^\circ$
Since, AOB is a straight line, we have: $\angle\text{AOE}+\angle\text{COE}+\angle\text{BOC}=180^\circ$
$\Rightarrow 3x + 5x + 2x = 180^\circ$
$\Rightarrow 10x = 180^\circ$
$\Rightarrow x = 18^\circ$
Therefore, $\angle\text{AOD}=2\times18^\circ=36^\circ$
$\angle\text{COE}=5\times18^\circ=90^\circ$
$\angle\text{AOE}=3\times18^\circ=54^\circ$
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