Ta có
\(\Delta ABC=\Delta D\)EF
=> \(\widehat{B}=\widehat{E}=55^0\)
Mà \(\widehat{A}+\widehat{B}=130^0\Rightarrow\widehat{A}=130^0-55^0=75^0\)
Lại có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Rightarrow\widehat{C}=180^0-55-75=50^0\)
Vậy
\(\widehat{A}=\widehat{D}=75^0\)
\(\widehat{B}=\stackrel\frown{E}=55^0\)
\(\widehat{C}=\widehat{F}=50^0\)