\(\Delta ABC:\widehat{B}=50^o;\widehat{A}=70^o;\widehat{C}=180^o-\left(50^o+70^o\right)=60^o\\ Vậy:\widehat{B}< \widehat{C}< \widehat{A}\Rightarrow AC< AB< BC\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{C}=60^o\)
\(\Rightarrow\widehat{\text{A}\text{ }}>\widehat{C}>\widehat{B}\)
\(\Rightarrow BC>AB>AC\)
Ta có: \(\widehat{A}+\)\(\widehat{B}+\)\(\widehat{C}=\)\(180^o\)\(\Rightarrow\widehat{C}\)=\(180^o\)\(-\widehat{A}\)\(-\widehat{B}\)=\(60^o\)
\(\Rightarrow\widehat{A}\)\(>\)\(\widehat{C}>\)\(\widehat{B}\Rightarrow\)\(BC>\)\(AB>\)\(AC\)
