In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°.

From the given right-angled triangle, we have:

$\frac{B C}{A B}=\tan 30^{\circ}$

$\Rightarrow \frac{6}{A B}=\frac{1}{\sqrt{3}}$

$\Rightarrow A B=6 \sqrt{3} \mathrm{~cm}$

Also, $\frac{B C}{A C}=\sin 30^{\circ}$

$\Rightarrow \frac{6}{A C}=\frac{1}{2}$

$\Rightarrow A C=(2 \times 6)=12 \mathrm{~cm}$

$\therefore A B=6 \sqrt{3} \mathrm{~cm}$ and $A C=12 \mathrm{~cm}$