\(\widehat{A}=\widehat{D}=32^0\)
\(\widehat{B}=\widehat{E}=78^0\)
\(\widehat{C}=\widehat{F}=70^0\)
Xét ΔABC ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\\
\Rightarrow32^o+78^o+\widehat{C}=180^o\\
\Rightarrow\widehat{C}=70^o\)
Vì ΔABC = ΔDEF
\(\Rightarrow\widehat{A}=\widehat{D}=32^o\) (2 góc tương ứng)
\(\widehat{B}=\widehat{E}=78^o\) (2 góc tương ứng)
\(\widehat{C}=\widehat{F}=70^o\) (2 góc tương ứng)
\(\Delta ABC=\Delta DEF=>\widehat{A}=\widehat{D};\widehat{B}=\widehat{E};\widehat{C}=\widehat{F}\)
\(\widehat{D}=32^0;\widehat{E}=78^0\)
Mà \(\widehat{D}+\widehat{E}+\widehat{F}=180^0\)
\(=>\widehat{F}=180^0-\left(32^0+78^0\right)\)
\(=>\widehat{F}=70^0\)
Vậy \(\widehat{D}=32^0;\widehat{E}=78^0\);\(\widehat{F}=70^0\)