Ta có:
\(\widehat{A}=5\widehat{C}\Rightarrow\widehat{C}=\frac{\widehat{A}}{5}\left(1\right)\)
\(\widehat{B}=3\widehat{C}\Rightarrow\widehat{C}=\frac{\widehat{B}}{3}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\frac{\widehat{C}}{1}=\frac{\widehat{A}}{5}=\frac{\widehat{B}}{3}=\frac{\widehat{C}+\widehat{A}+\widehat{B}}{1+5+3}=\frac{180^0}{9}=20^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=20^0.5=100^0\\\widehat{B}=20^0.3=60^0\\\widehat{C}=20^0.1=20^0\end{matrix}\right.\)
Ta có: \(\widehat{A}=5.\widehat{C}\Leftrightarrow\frac{\widehat{A}}{5}=\widehat{C};\widehat{B}=3.\widehat{C}\Leftrightarrow\frac{\widehat{B}}{3}=\widehat{C}\)
\(\Rightarrow\frac{\widehat{A}}{5}=\frac{\widehat{B}}{3}=\widehat{C};\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
Áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{\widehat{A}}{5}=\frac{\widehat{B}}{3}=\widehat{C}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{5+3+1}=\frac{180^o}{9}=20\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=100^o\\\widehat{B}=60^o\\\widehat{C}=20^o\end{matrix}\right.\)
