Question

Answer 1

Bài 2:

Xét tam giác ABC vuông tại A:

\(cosB=\dfrac{AB}{BC}=\dfrac{12}{20}=\dfrac{3}{5}\Rightarrow\widehat{B}\approx53^0\)

Bài 3:

Áp dụng HTL:

\(ED^2=EK.EF\Rightarrow EF=\dfrac{ED^2}{EK}=\dfrac{9^2}{5,4}=15\left(cm\right)\)

\(\left\{{}\begin{matrix}DK^2+EK^2=ED^2\\DF^2+DE^2=EF^2\end{matrix}\right.\)(Pytago)

\(\Rightarrow\left\{{}\begin{matrix}DK=\sqrt{ED^2-EK^2}=\sqrt{9^2-5,4^2}=7,2\left(cm\right)\\DF=\sqrt{EF^2-DE^2}=\sqrt{15^2-9^2}=12\left(cm\right)\end{matrix}\right.\)

Áp dụng tslg:

\(tanDEF=\dfrac{DF}{ED}=\dfrac{12}{9}=\dfrac{4}{3}\)