Question
In the given figure, three lines AB, CD and EF intersect at a point O such that $\angle\text{AOE}=35^\circ$ and $\angle\text{BOD}=40^\circ.$ Find the measure of $\angle\text{AOC},\angle\text{BOF},\angle\text{COF}$ and $\angle\text{DOE}.$
✓
Answer
In the given figure,$\angle\text{AOC}=\angle\text{BOD}=40^\circ$ (Vertically opposite angles)
$\angle\text{BOF}=\angle\text{AOE}=35^\circ$ (Vertically opposite angles)
Now, $\angle\text{EOC}$ and $\angle\text{COF}$ form a linear pair.$\therefore\angle\text{EOC}+\angle\text{COF}=180^\circ$
$\Rightarrow(\angle\text{AOE}+\angle\text{AOC})+\angle\text{COF}=180^\circ$
$\Rightarrow35^\circ+406\circ+\angle\text{COF}=180^\circ$
$\Rightarrow75^\circ+\angle\text{COF}=180^\circ$
$\Rightarrow\angle\text{COF}=180^\circ-75^\circ=105^\circ$
Also, $\angle\text{DOE}=\angle\text{COF}=105^\circ$ (Vertically opposite angles)
$\angle\text{BOF}=\angle\text{AOE}=35^\circ$ (Vertically opposite angles)
Now, $\angle\text{EOC}$ and $\angle\text{COF}$ form a linear pair.$\therefore\angle\text{EOC}+\angle\text{COF}=180^\circ$
$\Rightarrow(\angle\text{AOE}+\angle\text{AOC})+\angle\text{COF}=180^\circ$
$\Rightarrow35^\circ+406\circ+\angle\text{COF}=180^\circ$
$\Rightarrow75^\circ+\angle\text{COF}=180^\circ$
$\Rightarrow\angle\text{COF}=180^\circ-75^\circ=105^\circ$
Also, $\angle\text{DOE}=\angle\text{COF}=105^\circ$ (Vertically opposite angles)
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