a.   Kẻ $BK\perp AC$  =>  $\widehat{KBC}=60^{\circ}$

=>  $\widehat{KBA}=\widehat{KBC}-\widehat{ABC}=60^{\circ}-38^{\circ}=22^{\circ}$

Xét tam giác vuông KBC ( $\widehat{K}=90^{\circ}$ ), ta có :

$BK=BC.\sin C=11.\sin 30^{\circ}=5,5(cm)$

Xét tam giác vuông KBA ( $\widehat{K}=90^{\circ}$ ), ta có :

$AB=\frac{BK}{\cos 22^{\circ}}=\frac{5,5}{\cos 22^{\circ}}\approx 5,932(cm)$

Xét tam giác vuông ABN ( $\widehat{N}=90^{\circ}$ ), ta có :

$AN=AB.\sin 38^{\circ}\approx 5,932.\sin 38^{\circ}\approx 3,652(cm)$

Vậy $AN\approx 3,652(cm)$ .

b.  Xét tam giác vuông ANC ( $\widehat{N}=90^{\circ}$ ), ta có :

$AC=\frac{AN}{\sin C}\approx \frac{3,652}{\sin 30^{\circ}}\approx 7,304(cm)$

Vậy  $AC\approx 7,304(cm)$ .