Xét ΔABC vuông tại A có\(\widehat{ABC}+\widehat{ACB}=90^0\)=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)=>\(\widehat{OBC}+\widehat{OCB}=45^0\)Xét ΔBOC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)=>\(\widehat{BOC}+45^0=180^0\)=>\(\widehat{BOC}=135^0\)Câu trả lời: Xét ΔABC vuông tại A có\(\widehat{ABC}+\widehat{ACB}=90^0\)=>\(2\cdot\left(\widehat{OBC}+\widehat{OCB}\right)=90^0\)=>\(\widehat{OBC}+\widehat{OCB}=45^0\)Xét ΔBOC có \(\widehat{BOC}+\widehat{OBC}+\widehat{OCB}=180^0\)=>\(\widehat{BOC}+45^0=180^0\)=>\(\widehat{BOC}=135^0\)