In $\triangle ABC$,right-angled at $B$,$AB = 5 \, cm$ and $\angle ACB = 30^{\circ}$. Determine the lengths of the sides $BC$ and $AC$.

(N/A) To find the length of the side $BC$,we will choose the trigonometric ratio involving $BC$ and the given side $AB$. Since $BC$ is the side adjacent to angle $C$ and $AB$ is the side opposite to angle $C$,therefore:
$\frac{AB}{BC} = \tan C$
$\frac{5}{BC} = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$
Which gives $BC = 5\sqrt{3} \, cm$.
To find the length of the side $AC$,we consider:
$\sin 30^{\circ} = \frac{AB}{AC}$
$\frac{1}{2} = \frac{5}{AC}$
$AC = 10 \, cm$.
Alternatively,we could have used the Pythagoras theorem to determine the third side:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + (5\sqrt{3})^2} \, cm = \sqrt{25 + 75} \, cm = \sqrt{100} \, cm = 10 \, cm$.