5. In the figure on the right, $|A C|=|B C|=1$ unit, $\alpha=30^{\circ}$, and $\angle A C B=90^{\circ}$. Find the area of $\triangle A D C$.
Answer: $\frac{1}{4}(3-\sqrt{3}) \quad$ (Please accept $\frac{1}{4}(12-6 \sqrt{3})$, in case they used half-angle formulas instead.)
Solution: Since $|A C|=|B C|$ and $\angle A C B=90^{\circ}$, it follows $\angle D A C=\angle C B D=45^{\circ}$. Because $\angle \alpha=30^{\circ}$ and $\angle A C D=60^{\circ}$, we have $\angle C D A=75^{\circ}$. Utilizing the Sine Law, we get
$$
\frac{|C D|}{\sin 45^{\circ}}=\frac{|A C|}{\sin 75^{\circ}}=\frac{1}{\sin 75^{\circ}} .
$$

This leads to
$$
|C D|=\frac{\sin 45^{\circ}}{\sin 75^{\circ}}=\sqrt{3}-1 .
$$

Then the altitude of $\triangle A D C$ with respect to the base $A C$ has length
$$
h=(\sqrt{3}-1) \times \sin 60^{\circ}=\frac{1}{2}(3-\sqrt{3}) .
$$

Therefore, the area of $\triangle A D C$ is $\frac{1}{4}(3-\sqrt{3})$ square units.