Question
Two lines $AB$ and $CD$ intersect each other at a point $O$ such that $\angle\text{AOC}:\angle\text{AOD}=5:7.$ Find all the angles. 
✓
Answer
Let $\angle\text{AOC}=5\text{k}$ and $\angle\text{AOD}=7\text{k,}$ where $k$ is some constant.
Here, $\angle\text{AOC}$ and $\angle\text{AOD}$ form a linear pair.
$\therefore\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow5\text{k}+7\text{k}=180^\circ$
$\Rightarrow12\text{k}=180^\circ$
$\Rightarrow\text{k}=15^\circ$
$\therefore\angle\text{AOC}=5\text{k}=5\times15^\circ=75^\circ$
$\angle\text{AOD}=7\text{k}=7\times15^\circ=105^\circ$
Now, $\angle\text{BOD}=\angle\text{AOC}=75^\circ$ (Vertically opposite angles)
$\angle\text{BOC}=\angle\text{AOD}=105^\circ$ (Vertically opposite angles)
Here, $\angle\text{AOC}$ and $\angle\text{AOD}$ form a linear pair.
$\therefore\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow5\text{k}+7\text{k}=180^\circ$
$\Rightarrow12\text{k}=180^\circ$
$\Rightarrow\text{k}=15^\circ$
$\therefore\angle\text{AOC}=5\text{k}=5\times15^\circ=75^\circ$
$\angle\text{AOD}=7\text{k}=7\times15^\circ=105^\circ$
Now, $\angle\text{BOD}=\angle\text{AOC}=75^\circ$ (Vertically opposite angles)
$\angle\text{BOC}=\angle\text{AOD}=105^\circ$ (Vertically opposite angles)
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