Question
In figure, lines $AB, CD$ and $EF$ intersect at $O$. Find the measures of $\angle\text{AOC},\angle\text{COF},\angle\text{DOE}$ and $\angle\text{BOF}.$
✓
Answer
$\angle\text{AOE}+\angle\text{DOE}+\angle\text{BOD}=180^\circ$ [linear pair]
$\Rightarrow\ \angle\text{DOE}=180^\circ-40^\circ-35^\circ=105^\circ$
$\angle\text{DOE}=\angle\text{COF}=105^\circ$ [vertically opposite angles]
Now, $\angle\text{AOE}+\angle\text{AOF}=180^\circ$ [lineat pair]
$\Rightarrow\ \angle\text{AOE}+\angle\text{AOC}+\angle\text{COF}=180^\circ$
$\Rightarrow\ 40^\circ+\angle\text{AOC}+105^\circ=180^\circ $
$\Rightarrow\ \angle\text{AOC}=35^\circ$
Also, $\angle\text{BOF}=\angle\text{AOE}=40^\circ$ [vertically opposite angles]
$\Rightarrow\ \angle\text{DOE}=180^\circ-40^\circ-35^\circ=105^\circ$
$\angle\text{DOE}=\angle\text{COF}=105^\circ$ [vertically opposite angles]
Now, $\angle\text{AOE}+\angle\text{AOF}=180^\circ$ [lineat pair]
$\Rightarrow\ \angle\text{AOE}+\angle\text{AOC}+\angle\text{COF}=180^\circ$
$\Rightarrow\ 40^\circ+\angle\text{AOC}+105^\circ=180^\circ $
$\Rightarrow\ \angle\text{AOC}=35^\circ$
Also, $\angle\text{BOF}=\angle\text{AOE}=40^\circ$ [vertically opposite angles]
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