Sửa: $\widehat{AOD}=70^\circ$
Ta có: $AB∩CD≡O$
$→\widehat{AOD}$ đối đỉnh $\widehat{BOC}$
$→\widehat{AOD}=\widehat{BOC}=70^\circ$
Ta có: $\widehat{AOD},\widehat{AOC}$ là 2 góc kề bù
$→\widehat{AOD}+\widehat{AOC}=180^\circ$
hay $70^\circ+\widehat{AOC}=180^\circ$
$→\widehat{AOC}=180^\circ-70^\circ=110^\circ$
Vì $AB∩CD≡O$
$→\widehat{AOC}$ đối đỉnh $\widehat{BOD}$
$→\widehat{AOC}=\widehat{BOD}=110^\circ$
Vậy $\widehat{BOC}=70^\circ,\widehat{AOC}=\widehat{BOD}=110^\circ$