The slope of our roof is $30^{\circ}$ towards the south and $15^{\circ}$ towards the west. What angle does the horizontal edge of the roof form with the north direction?
Let's take a point $P$ on the upper edge of the roof, and project it onto the horizontal plane. Place a compass at the projection point $P^{\prime}$, and mark the point $D$ corresponding to the southern direction and the point $N$ corresponding to the western direction on the lower horizontal edge of the roof, such that $D P^{\prime} N \varangle=90^{\circ}$. The slope of the roof in the southern direction is measured by the angle between the lines $P D$ and $D P^{\prime}$, that is, $P D P^{\prime} \varangle=30^{\circ}$, and similarly, $P N P^{\prime} \varangle=15^{\circ}$. The angle between the horizontal edge and the north direction is $\alpha$, which is $N D P^{\prime} \varangle$.

Since $P D P^{\prime}$ is half of an equilateral triangle with side length 2 units, its height is $P^{\prime} D=2 \frac{\sqrt{3}}{2}=\sqrt{3}$. From the right triangle $P P^{\prime} N$, we have $N P^{\prime}=\frac{1}{\operatorname{tg} 15^{\circ}} \cdot \mathrm{In}$ the right triangle $N P^{\prime} D$,

$$
\operatorname{tg} \alpha=\frac{N P^{\prime}}{P^{\prime} D}=\frac{\frac{1}{\operatorname{tg} 15^{\circ}}}{\sqrt{3}} \approx 2.1547, \quad \text { hence } \alpha \approx 65.1^{\circ}
$$