Áp dụng tc dtsbn:
\(\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+4}=\dfrac{180^0}{9}=20^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A}=40^0\\\widehat{B}=60^0\\\widehat{C}=80^0\end{matrix}\right.\)

