Bài 2:
a: \(\hat{BAE}\) là góc ngoài tại đỉnh A của ΔABC
=>\(\hat{BAE}=\frac{180^0-\hat{BAC}}{2}=\frac{\hat{ABC}+\hat{ACB}}{2}\)
Ta có: \(\hat{ABE}+\hat{ABC}=180^0\)
=>\(\hat{ABE}=180^0-\hat{ABC}\)
Xét ΔAEB có \(\hat{ABE}+\hat{BAE}+\hat{AEB}=180^0\)
=>\(\hat{AEB}=180^0-180^0+\hat{ABC}-\frac{\hat{ABC}+\hat{ACB}}{2}=\frac{2\cdot\hat{ABC}-\hat{ABC}-\hat{ACB}}{2}=\frac{\hat{ABC}-\hat{ACB}}{2}\)
b: \(\hat{AEB}=\frac12\left(\hat{ABC}-\hat{ACB}\right)\)
=>\(\hat{ABC}-\hat{ACB}=2\cdot\hat{AEB}=2\cdot15^0=30^0\)
Xét ΔABC có \(\hat{ABC}+\hat{ACB}+\hat{BAC}=180^0\)
=>\(\hat{ABC}+\hat{ACB}=180^0-60^0=120^0\)
mà \(\hat{ABC}-\hat{ACB}=30^0\)
nên \(\hat{ABC}=\frac{120^0+30^0}{2}=75^0;\hat{ACB}=75^0-30^0=45^0\)
