a, Vì \(\widehat{AOB}< \widehat{AOC}\left(40^o< 70^o\right)\) nên tia OB nằm giữa 2 tia OA và OC
=> \(\widehat{AOB}+\widehat{BOC}=\widehat{AOC}\)
\(40^o+\widehat{BOC}=70^o\)
\(\widehat{BOC}=70^o-40^o=30^o\)
b, Ta có: \(\widehat{AOB}+\widehat{AOD}=180^o\) (kề bù)
\(40^o+\widehat{AOD}=180^o\)
\(\widehat{AOD}=180^o-40^o=140^o\)
Lại có: \(\widehat{BOC}+\widehat{COD}=180^o\) (kề bù)
\(30^o+\widehat{COD}=180^o\)
\(\widehat{COD}=180^o-30^o=150^o\)