Ta có: \(\left\{{}\begin{matrix}\widehat{A}=2\widehat{B}\\\widehat{B}=3\widehat{C}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{1}\\\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{1}\end{matrix}\right.\)\(\Rightarrow\dfrac{\widehat{A}}{6}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{1}\)
Áp dụng t/c dtsbn:
\(\dfrac{\widehat{A}}{6}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{1}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{6+3+1}=\dfrac{180^0}{10}=18^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=18^0.6=108^0\\\widehat{B}=18^0.3=54^0\\\widehat{C}=18^0.1=18^0\end{matrix}\right.\)
Áp dụng tc dtsnb:
\(\widehat{A}=2\widehat{B}=6\widehat{C}\Rightarrow\dfrac{\widehat{A}}{6}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{1}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{6+3+1}=\dfrac{180^0}{10}=18^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A}=108^0\\\widehat{B}=54^0\\\widehat{C}=18^0\end{matrix}\right.\)
