Ta có:A^-B^=18,B^-C^=18
\(\Rightarrow\)A^=18+B(1),B=18+C^
Thay B^ vào (1) ta có:
A^=18+18+C^
A^=36+C^
Mà A^+B^+C^=180(tổng 3 góc trong tam giác)
Hay 36+C^+18+C^+C^=180
3C^+54=180
3C^=126
C^=42
Vậy góc C=42 độ
Ta có: \(\widehat{A}-\widehat{B}=18\)*; \(\widehat{B}-\widehat{C}=18\)* (1)
\(\Rightarrow\widehat{A}-\widehat{B}=\widehat{B}-\widehat{C}=18\)*
\(\Rightarrow\) \(\widehat{A}-\widehat{B}-\widehat{B}+\widehat{C}=0\)*
\(\Rightarrow\widehat{A}-2\widehat{B}+\widehat{C}=0\)*
\(\Rightarrow\widehat{A}+\widehat{C}-2\widehat{B}=0\)*
\(\Rightarrow\widehat{A}+\widehat{C}=2\widehat{B}\)
Mà \(\widehat{A}+\widehat{B}+\widehat{C}=\left(\widehat{A}+\widehat{C}\right)+\widehat{B}=2\widehat{B}+\widehat{B}=3\widehat{B}=180\)*
Suy ra: \(\widehat{B}=\frac{180}{3}=60\)*
Thế \(\widehat{B}\) vào (1) ta được: \(60-\widehat{C}=18\)*
\(\Rightarrow\widehat{C}=60-18=42\)*
^-^
