Bài 1:
\(DF=\sqrt{EF^2-DE^2}=5\left(cm\right)\left(pytago\right)\\ \sin\widehat{E}=\cos\widehat{F}=\dfrac{DF}{EF}=\dfrac{5}{13}\\ \cos\widehat{E}=\sin\widehat{F}=\dfrac{DE}{EF}=\dfrac{12}{13}\\ \tan\widehat{E}=\cot\widehat{F}=\dfrac{DF}{DE}=\dfrac{5}{12}\\ \cot\widehat{E}=\tan\widehat{F}=\dfrac{DE}{DF}=\dfrac{12}{5}\)

