a,
$\Delta ABC$ vuông tại $A$, $AH\bot BC$ có:
$AH^2=HB.HC$
$\Leftrightarrow HC=6,4(cm)$
$BC=HC+HB=10(cm)$
$AB^2=HB.BC$
$\Rightarrow AB=\sqrt{10.3,6}=6(cm)$
$\Rightarrow AC=\sqrt{BC^2-AB^2}=8(cm)$
b,
$\sin B=\dfrac{AB}{BC}=\dfrac{6}{10}=0,6$
$\Rightarrow \widehat{B}\approx 36^o52'$
$\Rightarrow \widehat{C}=90^o-\widehat{B}=53^o8'$