Xét ΔDEF có
\(\widehat{D}+\widehat{E}+\widehat{F}=180^0\)(Định lí tổng ba góc trong một tam giác)
\(\Leftrightarrow\widehat{E}+\widehat{F}=150^0\)
\(\Leftrightarrow\dfrac{1}{2}\cdot\widehat{F}+\widehat{F}=150^0\)
\(\Leftrightarrow\dfrac{3}{2}\cdot\widehat{F}=150^0\)
hay \(\widehat{F}=100^0\)
Vì \(\widehat{E}+\widehat{F}=150^0\)
nên \(\widehat{E}+100^0=150^0\)
hay \(\widehat{E}=50^0\)
Vậy: \(\widehat{F}=100^0\); \(\widehat{E}=50^0\)