a/ Ta có:
\(\hat{AOC}=\hat{AOB}+\hat{BOC}\)
\(hay:120\text{°}=60\text{°}+\hat{BOC}\)
\(\Rightarrow\hat{BOC}=120\text{°}-60\text{°}=60\text{°}\)
Vậy: \(\hat{BOC}=60\text{°}\)
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b/ Ta có:
\(\hat{BOC}=60\text{°}\) ; \(\hat{AOB}=60\text{}\text{°}\)
Vậy: Tia OB là tia phân giác của \(\hat{AOC}\)
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c/ \(\hat{BOC}=60\text{°}\)
\(\hat{COD}+\hat{AOC}=180\text{}\text{}\text{}\text{°}\) (kề bù)
hay: \(\hat{COD}+120\text{°}=180\text{°}\)
\(\Rightarrow\hat{COD}=180\text{°}-120\text{°}=60\text{°}\)
- OM là tia phân giác của \(\hat{COD}\) (gt)
\(\Rightarrow\hat{COM}=\dfrac{\hat{COD}}{2}=\dfrac{60\text{°}}{2}=30\text{°}\)
Suy ra: \(\hat{BOC}+\hat{COM}=60\text{°}+30\text{°}=90\text{°}\)
Vậy: \(\hat{BOC}\&\hat{COM}\) phụ nhau (đpcm).
