\(a,DE=\sqrt{DF^2-EF^2}=16\left(cm\right)\left(pytago\right)\)
\(b,\) Áp dụng HTL: \(\left\{{}\begin{matrix}FK=\dfrac{EF^2}{DF}=7,2\left(cm\right)\\KE=\dfrac{EF\cdot ED}{DF}=9,6\left(cm\right)\\EF=12\left(cm\right)\end{matrix}\right.\)
\(\sin\widehat{F}=\dfrac{EK}{EF}=\dfrac{9,6}{12}=\dfrac{4}{5}\approx\sin53^0\\ \Rightarrow\widehat{F}\approx53^0\\ \Rightarrow\widehat{FEK}=90^0-\widehat{F}\approx37^0\)
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