Cho \(\Delta ABC=\Delta DEF\). Biết \(\widehat{A}+\widehat{B}=130^0\), \(\widehat{E}=55^0\). Tính \(\widehat{A}\), \(\widehat{C}\), \(\widehat{D}\), \(\widehat{F}\)?
\(\widehat{A}=\widehat{D}=65^0\), \(\widehat{C}=\widehat{F}=50^0\).\(\widehat{A}=\widehat{D}=50^0\), \(\widehat{C}=\widehat{F}=65^0\).\(\widehat{A}=\widehat{D}=75^0\), \(\widehat{C}=\widehat{F}=50^0\).\(\widehat{A}=\widehat{D}=50^0\), \(\widehat{C}=\widehat{F}=75^0\).Hướng dẫn giải:Vì \(\Delta ABC=\Delta DEF\) nên \(\widehat{A}=\widehat{D}\), \(\widehat{B}=\widehat{E}=55^0\), \(\widehat{C}=\widehat{F}\)
Xét tam giác \(ABC\) có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) mà \(\widehat{A}+\widehat{B}=130^0\) \(\Rightarrow\)\(\widehat{C}=50^0\) \(\Rightarrow\) \(\widehat{C}=\widehat{F}=50^0\)
Lại có \(\widehat{A}+\widehat{B}=130^0\) mà \(\widehat{B}=\widehat{E}=55^0\) \(\Rightarrow\) \(\widehat{A}=\widehat{D}=75^0\)
