Question
Two lines AB and CD intersect at O. If $\angle\text{AOC}=50^\circ,$ find $\angle\text{AOD},\angle\text{BOD}$ and $\angle\text{BOC}.$
✓
Answer
We know that if two lines intersect then the vertically-opposite angle are equal.
Therefore,
$\angle\text{AOC}=\angle\text{BOD}=50^\circ$
Let $\angle\text{AOD}=\angle\text{BOC}=\text{x}^\circ$
Also, we know that the sum of all angles around a point is 360°.
Therefore,
$\angle\text{AOC}+\angle\text{AOD}+\angle\text{BOD}+\angle\text{BOC}=360^\circ$
⇒ 50 + x + 50 + x = 360º
⇒ 2x = 260º
⇒ x = 130º
Hence, $\angle\text{AOD}=\angle\text{BOC}=130^\circ$
Therefore, $\angle\text{AOD}=130^\circ,\angle\text{BOD}=50^\circ$ and $\angle\text{BOC}=130^\circ.$
Therefore,
$\angle\text{AOC}=\angle\text{BOD}=50^\circ$
Let $\angle\text{AOD}=\angle\text{BOC}=\text{x}^\circ$
Also, we know that the sum of all angles around a point is 360°.
Therefore,
$\angle\text{AOC}+\angle\text{AOD}+\angle\text{BOD}+\angle\text{BOC}=360^\circ$
⇒ 50 + x + 50 + x = 360º
⇒ 2x = 260º
⇒ x = 130º
Hence, $\angle\text{AOD}=\angle\text{BOC}=130^\circ$
Therefore, $\angle\text{AOD}=130^\circ,\angle\text{BOD}=50^\circ$ and $\angle\text{BOC}=130^\circ.$
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