Xét ΔABC có \(\hat{BAC}+\hat{ABC}+\hat{ACB}=180^0\)
=>\(\hat{ABC}+\hat{ACB}=180^0-40^0=140^0\)
=>\(\hat{ABO}+\hat{ACO}+\hat{OBC}+\hat{OCB}=140^0\)
=>\(\hat{OBC}+\hat{OCB}+30^0+45^0=140^0\)
=>\(\hat{OBC}+\hat{OCB}=140^0-30^0-45^0=140^0-75^0=65^0\)
Xét ΔOBC có \(\hat{OBC}+\hat{OCB}+\hat{BOC}=180^0\)
=>\(\hat{BOC}=180^0-65^0=115^0\)
=>\(x=115^0\)

