Có: `hat{AOC}` là góc ở tâm chắn $\mathop{AC}\limits^{\displaystyle\frown}$

nên sđ $\mathop{AC}\limits^{\displaystyle\frown}$ `= 135^o`

Có: $\triangle$`OAC` cân tại `O` `(OA = OC = R)`

nên `hat{OAC} = hat{OCA} = (180^o - 135^o)/2 = 22,5^o`

Có: `hat{BAC} = hat{BAO} + hat{OAC}` (góc kề)

`hat{BAC} = 60^o + 22,5^o = 82,5^o`

Có: `hat{BAC}` là góc nội tiếp chắn $\mathop{BC}\limits^{\displaystyle\frown}$

nên sđ $\mathop{BC}\limits^{\displaystyle\frown}$ `= 2 hat{BAC} = 2 . 82,5^o = 165^o`

Có: sđ $\mathop{AB}\limits^{\displaystyle\frown}$ `+` sđ $\mathop{AC}\limits^{\displaystyle\frown}$ `+` $\mathop{BC}\limits^{\displaystyle\frown}$ `= 360^o`

$\mathop{AB}\limits^{\displaystyle\frown}$ `= 360^o -` $\mathop{AC}\limits^{\displaystyle\frown}$ - $\mathop{BC}\limits^{\displaystyle\frown}$ `= 360^o - 135^o - 165^o = 60^o`