Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) \(^{\left(1\right)}\)
Thay \(\widehat{A}=5\widehat{C},\widehat{B}=3\widehat{C}\) vào \(^{\left(1\right)}\), ta được:
\(5\widehat{C}+3\widehat{C}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{C}=20^o\)
\(\Rightarrow\widehat{A}=5.20^o=100^o,\widehat{B}=3.20^o=60^o\)
Tam giác $ABC$ có: \(\widehat{A}=5\widehat{C};\widehat{B}=3\widehat{C}\)
Mà ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow5\widehat{C}+3\widehat{C}+\widehat{C}=180^o\\ \Rightarrow9\widehat{C}=180^o\\ \Rightarrow\widehat{C}=20^o\)
\(\cdot\widehat{A}=5\widehat{C}=5.20^o=100^o\\ \cdot\widehat{B}=3\widehat{C}=3.20^o=60^o\)
