Giải thích các bước giải:
a.Ta có :
$\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^o$
$\to 2\widehat{ACB}+\widehat{ACB}+60^o=180^o$
$\to \widehat{ACB}=40^o$
b.Từ câu a
$\to \widehat{ABC}=180^o-\widehat{ACB}-\widehat{BAC}=80^o$
$\to \widehat{IBC}=\dfrac 12\widehat{ABC}=40^o,\widehat{ICB}=\dfrac 12\widehat{ACB}=20^o$
$\to\widehat{BIC}=180^o-\widehat{IBC}-\widehat{ICB}=120^o$