Ta có $\Delta OBC$ là tam giác đều (OB=OC=BC=R)

Suy ra: $\widehat{BOC}={{60}^{0}}$ $\Rightarrow sd\overset\frown{BC}={{60}^{0}}$ $\Rightarrow \widehat{ABC}=\frac{1}{2}sd\overset\frown{BC}=\frac{1}{2}{{.60}^{0}}={{30}^{0}}$

Trong tứ giác OBAC có: $\widehat{A}+\widehat{B}+\widehat{C}+\widehat{O}={{360}^{0}}\Leftrightarrow \widehat{A}={{360}^{0}}-(\widehat{B}+\widehat{C}+\widehat{O})={{360}^{0}}-(\widehat{{{90}^{0}}}+\widehat{{{90}^{0}}}+\widehat{{{60}^{0}}})={{120}^{0}}$