Answer only TWO questions from this Section. Please, note that if you answer more than two questions, only the first two will be marked.

The sum of angles in a triangle is $180^\circ$.

$$\angle BAC = 180^\circ - \angle ABC - \angle ACB$$ $$\angle BAC = 180^\circ - 59^\circ - 73^\circ$$ $$\angle BAC = 180^\circ - 132^\circ$$ $$\angle BAC = 48^\circ$$

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
We have $|BC| = 12$ cm, $\angle BAC = 48^\circ$, $\angle ABC = 59^\circ$, and $\angle ACB = 73^\circ$.
To find $|AC|$, we use the ratio:

$$\frac{|AC|}{\sin(\angle ABC)} = \frac{|BC|}{\sin(\angle BAC)}$$ $$|AC| = \frac{|BC| \times \sin(\angle ABC)}{\sin(\angle BAC)}$$ $$|AC| = \frac{12 \times \sin 59^\circ}{\sin 48^\circ}$$ $$|AC| = \frac{12 \times 0.857167}{0.743145}$$ $$|AC| \approx 13.84 \text{ cm}$$

To find $|AB|$, we use the ratio:

$$\frac{|AB|}{\sin(\angle ACB)} = \frac{|BC|}{\sin(\angle BAC)}$$ $$|AB| = \frac{|BC| \times \sin(\angle ACB)}{\sin(\angle BAC)}$$ $$|AB| = \frac{12 \times \sin 73^\circ}{\sin 48^\circ}$$ $$|AB| = \frac{12 \times 0.956305}{0.743145}$$ $$|AB| \approx 15.44 \text{ cm}$$

The lengths of the remaining two sides are:
$|AC| \approx \boxed{\text{13.84 cm}}$
$|AB| \approx \boxed{\text{15.44 cm}}$