Bài giải:

Xét tam giác ABC ta có:

$\widehat{BAC}+\widehat{ABC}+\widehat{ACB}={{180}^{0}}$ (tổng ba góc trong tam giác)

$\Rightarrow \widehat{ABC}+\widehat{ACB}={{180}^{0}}-\widehat{BAC}={{180}^{0}}-{{60}^{0}}={{120}^{0}}$

$\Rightarrow \frac{1}{2}\widehat{ABC}+\frac{1}{2}\widehat{ACB}=\frac{1}{2}{{.120}^{0}}={{60}^{0}}$

Xét tam giác OBC ta có:

$\widehat{OBC}+\widehat{OCB}+\widehat{BOC}={{180}^{0}}$ (tổng ba góc trong tam giác)

$\Rightarrow \widehat{BOC}={{180}^{0}}-(\widehat{OBC}+\widehat{OCB})={{180}^{0}}-\left( \frac{1}{2}\widehat{ABC}+\frac{1}{2}\widehat{ACB} \right)={{180}^{0}}-{{60}^{0}}={{120}^{0}}$

Vậy đáp án đúng là: D