Ta có: \(\hat{NCB}+\hat{NCx}=180^0\) (hai góc kề bù)
=>\(\hat{ACB}=180^0-130^0=50^0\)
MN//BC
=>\(\hat{AMN}=\hat{ABC}\) (hai góc đồng vị)
=>\(\hat{ABC}=45^0\)
Xét ΔABC có \(\hat{ABC}+\hat{ACB}+\hat{BAC}=180^0\)
=>\(\hat{BAC}=180^0-50^0-45^0=130^0-45^0=85^0\)
