Giải thích các bước giải:
a.Ta có $\widehat{AOC}+\widehat{BOC}=\widehat{AOB}=180^o$
$\to 2\widehat{BOC}+\widehat{BOC}=180^o$ vì $\widehat{AOC}=2\widehat{BOC}$
$\to 3\widehat{BOC}=180^o$
$\to \widehat{BOC}=60^o$
$\to \widehat{AOC}=120^o$
b.Ta có $Om, On$ là phân giác $\widehat{AOC},\widehat{BOC}$
$\to \widehat{mOn}=\widehat{mOC}+\widehat{COn}=\dfrac12\widehat{AOC}+\dfrac12\widehat{COB}=\dfrac12\widehat{AOB}=90^o$
