a)
$\widehat{B}+\widehat{C}=90{}^\circ \Rightarrow \widehat{C}=90{}^\circ -\widehat{B}=90{}^\circ -60{}^\circ =30{}^\circ $
$\cos B=\dfrac{AB}{BC}\Rightarrow AB=BC.\cos B=8.\cos 60{}^\circ =4cm$
$\sin B=\dfrac{AC}{BC}\Rightarrow AC=BC.\sin B=8.\sin 60{}^\circ =4\sqrt{3}cm$
b)
$\begin{cases}\cos B=\dfrac{HB}{AB}\Rightarrow AB.\cos B=HB\\\cos C=\dfrac{HC}{AC}\Rightarrow AC.\cos C=HC\end{cases}$
$\Rightarrow AB.\cos B+AC.\cos C=HB+HC=BC$
