Question
Two lines AB and CD intersect at a point O, such that $\angle\text{BOC}+\angle\text{AOD}=280^\circ,$ as shown in the figure. Find all the four angles.
✓
Answer
We know that if two lines intersect, then the vertically-opposite angles are equal. Let $\angle\text{BOC}=\angle\text{AOD}=\text{x}^\circ$ Then,$\text{x}+\text{x}=280$
$\Rightarrow2\text{x}=280$
$\Rightarrow\text{x}=140^\circ$
$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$
Also, let $\angle\text{AOC}=\angle\text{BOD}=\text{y}^\circ$ We know that the sum of all angles around a point is 360º.$\therefore\angle\text{AOC}+\angle\text{BOC}+\angle\text{BOD}+\angle\text{AOD}=360^\circ$
$\Rightarrow\text{y}+140+\text{y}+140=360^\circ$
$\Rightarrow2\text{y}=80^\circ$
$\text{y}=40^\circ$
Hence, $\angle\text{AOC}=\angle\text{BOD}=40^\circ$$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$ and $\angle\text{AOC}=\angle\text{BOD}=40^\circ$
$\Rightarrow2\text{x}=280$
$\Rightarrow\text{x}=140^\circ$
$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$
Also, let $\angle\text{AOC}=\angle\text{BOD}=\text{y}^\circ$ We know that the sum of all angles around a point is 360º.$\therefore\angle\text{AOC}+\angle\text{BOC}+\angle\text{BOD}+\angle\text{AOD}=360^\circ$
$\Rightarrow\text{y}+140+\text{y}+140=360^\circ$
$\Rightarrow2\text{y}=80^\circ$
$\text{y}=40^\circ$
Hence, $\angle\text{AOC}=\angle\text{BOD}=40^\circ$$\therefore\angle\text{BOC}=\angle\text{AOD}=140^\circ$ and $\angle\text{AOC}=\angle\text{BOD}=40^\circ$
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