Ta có: \(\widehat{AOB}+\widehat{BOC}=180^0\)(hai góc kề bù)
hay \(\widehat{BOC}=180^0-70^0\)
⇔\(\widehat{BOC}=110^0\)
Vậy: \(\widehat{BOC}=110^0\)
Có góc AOB và BOC là 2 góc kề bù
⇒AÔB+BÔC=\(\text{180}^{0}\)
Hay \(\text{70}^{0}\)+BÔC=\(\text{180}{^0}\)
⇒BÔC=\(\text{180}^{0}-\text{70}^{0}=\text{110}^{0}\)
Vậy BÔC=\(\text{110}^{0}\)