\(\left\{{}\begin{matrix}\widehat{A}+\widehat{B}=120^0\\-2\widehat{A}+3\widehat{B}=10^0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{A}=70^0\\\widehat{B}=50^0\end{matrix}\right.\)
=> ^C = 1800 - ^A - ^B = 600
Ta có ^A < ^C < ^B => BC < AB < AC

