Ta có: \(\widehat{A}+2\widehat{B}=100^0\)
\(\Rightarrow\widehat{A}=100^0-2\widehat{B}.\)
Xét \(\Delta ABC\) có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (định lí tổng 3 góc trong một tam giác).
=> \(100^0-2\widehat{B}+\widehat{B}+\widehat{C}=180^0\)
=> \(-2\widehat{B}+\widehat{B}+\widehat{C}=180^0-100^0\)
=> \(-2\widehat{B}+\widehat{B}+\widehat{C}=80^0\)
=> \(-\widehat{B}+\widehat{C}=80^0\)
=> \(\widehat{C}-\widehat{B}=80^0.\)
Vậy \(\widehat{C}-\widehat{B}=80^0.\)
Chúc bạn học tốt!
Có: \(\widehat{A}+2\widehat{B}=100^o\)
\(\Rightarrow\widehat{A}=100^o-2\widehat{B}\)
Xét △ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)(định lí tổng 3 góc của 1 tam giác)
\(\Rightarrow\left(100^o-2\widehat{B}\right)+\widehat{B}+\widehat{C}=180^o\\ 100^o-2\widehat{B}+\widehat{B}+\widehat{C}=180^o\\ -2\widehat{B}+\widehat{B}+\widehat{C}=100^o-180^o\\ -\widehat{B}+\widehat{C}=-80^o\\ \Leftrightarrow\widehat{C}-\widehat{B}=-80^o\)
