Ta có \(\widehat{A}-\widehat{D}=20^o\); \(\widehat{A}+\widehat{D}=180^o\)
Từ \(\widehat{A}-\widehat{D}=20^o\)
\(\Rightarrow\widehat{A}=20^o+\widehat{D}\)
Nên \(\widehat{A}+\widehat{D}=20^o+\widehat{D}+\widehat{D}=20^o+2\widehat{D}=180^o\)
\(\Rightarrow2\widehat{D}=160^o\Rightarrow\widehat{D}=80^o\)
Thay \(\widehat{D}=80^o\) vào \(\widehat{A}=20^o+\widehat{D}\) ta được \(\widehat{A}=20^o+80^o=100^o\)
Lại có \(\widehat{B}=2\widehat{C}\) ; \(\widehat{B}+\widehat{C}=180^o\)
nên \(2\widehat{C}+\widehat{C}=180^o\)
hay \(3\widehat{C}=180^o\) => \(\widehat{C}=60^o\)
Do đó \(\widehat{B}=2\widehat{C}=2.60^o\)
=> \(\widehat{B}=120^o.\)
