Ta có: $\widehat{C}=90^{\circ}-\widehat{B}=90^{\circ}-58^{\circ}=32^{\circ}$
=> $b = BC.\sin 58^{\circ} = a.\sin 58^{\circ} = 61,06 (cm)$
$c = BC.\cos 58^{\circ} = a.\cos 58^{\circ} = 38,15 (cm)$
=> $h_{a}=\frac{AB.AC}{BC}=\frac{c.b}{a}=\frac{38,15.61,06}{72}=32,35 (cm)$