Gọi \(\widehat{CBD}\) là góc ngoài của \(\widehat{ABC}\)
Theo định nghĩa góc ngoài, ta được
\(\widehat{ABC}+\widehat{DBC}=180^0\)(hai góc kề bù)
mà \(\widehat{DBC}=\frac{4}{3}\cdot\widehat{ABC}\), nên ta có:
\(\widehat{ABC}+\frac{4}{3}\cdot\widehat{ABC}=180^0\)
hay \(\widehat{ABC}\cdot\left(1+\frac{4}{3}\right)=180^0\)
\(\Leftrightarrow\widehat{ABC}=180^0:\frac{7}{3}=180^0\cdot\frac{3}{7}=\frac{540}{7}^0\)
Ta có: \(\Delta\)ABC vuông tại A(gt)
⇔\(\widehat{ABC}+\widehat{ACB}=90^0\)
hay \(\frac{540}{7}^0+\widehat{ACB}=90^0\)
⇔\(\widehat{ACB}=90^0-\frac{540}{7}^0=\frac{90}{7}^0\)
Vậy: \(\widehat{ABC}=\frac{540}{7}^0\); \(\widehat{ACB}=\frac{90}{7}^0\)