`a,` $\widehat{ACB}$=`180^o-60^o-67^o` = `53^o`

⇒$\mathop{AB}\limits^{\displaystyle\frown}$ = `2` · $\widehat{ACB}$ = `106^o`

⇒$\mathop{BC}\limits^{\displaystyle\frown}$ = `2` · $\widehat{BAC}$ = `134^o`

⇒$\mathop{CA}\limits^{\displaystyle\frown}$ = `2` · $\widehat{ABC}$ = `120^o`

`b,` Ta có: $\mathop{CA}\limits^{\displaystyle\frown}$ = `135^o`

⇒$\widehat{OAC}$=$\frac{180^o-135^o}{2}$ = `22^o30'`  (tam giác `OAC` cân `O`)

⇒$\widehat{BAC}$=$\widehat{BAO}$+$\widehat{OAC}$= `82^o30'`

⇔$\mathop{BC}\limits^{\displaystyle\frown}$ = `2` · $\widehat{BAC}$ = `165^o`

⇔$\mathop{AB}\limits^{\displaystyle\frown}$=`360^o -`$\mathop{BC}\limits^{\displaystyle\frown}$`-`$\mathop{CA}\limits^{\displaystyle\frown}$=`60^o`