a,
Ta có: $AB^2+AC^2=48^2+14^2=2500=50^2=BC^2$
$\to\Delta ABC$ vuông tại $A$, theo Pytago đảo
b,
$\widehat{A}=90^o$
$\sin B=\dfrac{AC}{BC}=\dfrac{7}{25}$
$\to \widehat{B}\approx 16^o15'$
$\to \widehat{C}=90^o-\widehat{B}\approx 73^o45'$
c,
$\Delta ABC$ có $CD$ phân giác
$\to \dfrac{AD}{BD}=\dfrac{AC}{BC}=\dfrac{7}{25}$
Đặt $AD=7x; BD=25x(cm)$ ($x>0$)
$\to 7x+25x=48$
$\to x=1,5$
$\to AD=10,5(cm); BD=37,5(cm)$
$CD=\sqrt{AD^2+AC^2}=17,5(cm)$