a/ Xét $ΔABC$ vuông tại $A$:
$·\,\,\cos C=\dfrac{AC}{BC}$ hay $\cos 30^\circ=\dfrac{AC}{4}$
$↔\dfrac{\sqrt 3}{2}=\dfrac{AC}{4}\\↔AC=2\sqrt 3(cm)$
$·\,\,\sin C=\dfrac{AB}{BC}$ hay $\sin 30^\circ=\dfrac{AB}{4}$
$↔\dfrac{1}{2}=\dfrac{AB}{4}\\↔AB=2(cm)$
Vậy $AC=2\sqrt 3cm,\,AB=2cm$
b/ Xét $ΔABC$:
$AD$ là đường phân giác $\widehat A$
$→\dfrac{AC}{AB}=\dfrac{DC}{DB}$ hay $\dfrac{2\sqrt 3}{2}=\dfrac{DC}{DB}$
$↔\sqrt 3=\dfrac{DC}{DB}\\↔DC=\sqrt 3 DB$
$DB+DC=BC\\↔DB+\sqrt 3DB=4\\↔(\sqrt 3+1)DB=4\\↔DB=\dfrac{4}{\sqrt 3+1}\\↔DB=2\sqrt 3-2(cm)\\→DC=BC-DB=4-2\sqrt 3+2=6-2\sqrt 3(cm)$
Vậy $DB=2\sqrt 3-2(cm),DC=6-2\sqrt 3(cm)$
