Đáp án:
$\begin{array}{l}
a)Trong:\Delta FEH:\widehat {FHE} = {90^0}\\
\Leftrightarrow \sin \widehat F = \dfrac{{EH}}{{FE}}\\
\Leftrightarrow \sin {40^0} = \dfrac{{EH}}{5}\\
\Leftrightarrow EH = \sin {40^0}.5 = 3,214\left( {cm} \right)\\
Vậy\,EH = 3,214cm\\
b)Trong:\Delta EHD:\widehat {EHD} = {90^0}\\
\Leftrightarrow \tan \widehat D = \dfrac{{EH}}{{HD}}\\
\Leftrightarrow HD = \dfrac{{EH}}{{\tan {{30}^0}}} = \dfrac{{3,214}}{{\dfrac{1}{{\sqrt 3 }}}} = 5,567\\
Theo\,Pytago:\\
E{D^2} = E{H^2} + H{D^2}\\
= {\left( {3,214} \right)^2} + 5,{567^2}\\
\Leftrightarrow ED = 6,428\left( {cm} \right)\\
Vậy\,HD = 5,567cm;ED = 6,428cm
\end{array}$
