Theo định lý tổng 3 góc trong tam giác \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\\ =>5\widehat{C}+60^o+\widehat{C}=180^o\\ =>6\widehat{C}=180^o-60^o\\ =>6\widehat{C}=120^o\\ =>\widehat{C}=120:6=20^o\)
Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(Tongbagoctrongmottamgiac\right)\)
\(\Rightarrow A+C=180^o-\widehat{B}\)
\(=180^o-60^o=120^o\)
Mà \(\widehat{A}=5\widehat{C}\)
\(\Rightarrow6\widehat{C}=120^o\)
\(\Rightarrow\widehat{C}=20^o\)