Bài làm :
Ta có :
`\hat{ABC}` `+` `\hat{ACB}` `=` $120^{o}$
⇒ $\dfrac{\hat{ABC} + \hat{ACB}}{2}$ `=` `\hat{IBC}` `+` `\hat{ICB}` `=` $51^{o}$
Tam giác IBC có :
⇒ `\hat{BIC}` = $180^{o}$ `- (` `\hat{IBC}` `+` `\hat{ICB}` `)` `=` $180^{o}$ `-` $51^{o}$ `=` $129^{o}$
Vậy `\hat{IBC}` `=` $129^{o}$