Question

Dùm trả lời ạ

Answer 1

\(\widehat{F}=90^0-\widehat{E}=30^0\)

\(tan\widehat{E}=\dfrac{DF}{DE}\Rightarrow DE=\dfrac{DF}{tan\widehat{E}}=\dfrac{10}{tan60^0}=\dfrac{10\sqrt{3}}{3}\left(cm\right)\)

\(sin\widehat{E}=\dfrac{DF}{EF}\Rightarrow EF=\dfrac{DF}{sin\widehat{E}}=\dfrac{10}{sin60^0}=\dfrac{20\sqrt{3}}{3}\left(cm\right)\)

Answer 2

Xét ΔDEF vuông tại D có 

\(\widehat{F}+\widehat{E}=90^0\)

hay \(\widehat{F}=30^0\)

Xét ΔDEF vuông tại D có 

\(\sin\widehat{F}=\dfrac{DE}{EF}\)

hay \(DE=10\sqrt{3}\left(cm\right)\)

\(\Leftrightarrow EF=20\sqrt{3}\left(cm\right)\)