Đặt \(\widehat{A}=2\widehat{B}=5\widehat{C}=k\)
\(\Rightarrow\hept{\begin{cases}\widehat{A}=k\\\widehat{B}=\frac{k}{2}\\\widehat{C}=\frac{k}{5}\end{cases}}\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=k+\frac{k}{2}+\frac{k}{5}=180^0\)
\(\Rightarrow\frac{17}{10}k=180^0\Leftrightarrow k=\frac{1800}{17}^0\)
\(\Rightarrow\hept{\begin{cases}\widehat{A}=\frac{1800}{17}^0\\\widehat{B}=\frac{900}{17}^0\\\widehat{C}=\frac{360}{17}^0\end{cases}}\)
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