Đáp án: $120^o$
Giải thích các bước giải:
Trường hợp $O$ nằm trong tam giác
Ta có $O$ là giao $3$ đường trung trực $\to OA=OB=OC$
$\to\Delta OAB,\Delta OBC,\Delta OAC$ cân tại $O$
$\to \widehat{OBC}+\widehat{OCB}=(\hat B-\widehat{OBA})+(\hat C-\widehat{OCA})$
$\to \widehat{OBC}+\widehat{OCB}=(\hat B-\widehat{OAB})+(\hat C-\widehat{OAC})$
$\to \widehat{OBC}+\widehat{OCB}=(\hat B+\hat C-\widehat{OAB}-\widehat{OAC})$
$\to \widehat{OBC}+\widehat{OCB}=(180^o-\hat A-(\widehat{OAB}+\widehat{OAC}))$
$\to \widehat{OBC}+\widehat{OCB}=(180^o-\hat A-\hat A)$
$\to \widehat{OBC}+\widehat{OCB}=180^o-2\hat A$
$\to \widehat{OBC}+\widehat{OCB}=60^o$
$\to 180^o-\widehat{BOC}=60^o$
$\to \widehat{BOC}=120^o$
Trường hợp $O$ nằm ngoài tam giác
Ta có:
$\widehat{OBC}+\widehat{OCB}=(\hat B-\widehat{OBA})+(\hat C+\widehat{OCA})$
$\to\widehat{OBC}+\widehat{OCB}=(\hat B+\hat C)+(\widehat{OCA}-\widehat{OBA})$
$\to\widehat{OBC}+\widehat{OCB}=(180^o-\hat A)-(\widehat{OAB}-\widehat{OAC})$
$\to\widehat{OBC}+\widehat{OCB}=(180^o-\hat A)-\hat A$
$\to\widehat{OBC}+\widehat{OCB}=180^o-2\hat A$
$\to \widehat{OBC}+\widehat{OCB}=60^o$
$\to 180^o-\widehat{BOC}=60^o$
$\to \widehat{BOC}=120^o$
