Answer:
$$
(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})=2-\sqrt{2}-\sqrt{3}+\sqrt{6}
$$

Consider a 360-sided regular polygon with side length 1 , rotated so that its sides are at half-degree inclinations (that is, its sides all have inclinations of $0.5^{\circ}, 1.5^{\circ}, 2.5^{\circ}$, and so on. Go to the bottom point on this polygon and then move clockwise, numbering the sides $1,2,3, \ldots, 360$ as you go. Then, take the section of 15 sides from side 31 to side 45 . These sides have inclinations of $30.5^{\circ}, 31.5^{\circ}, 32.5^{\circ}$, and so on, up to $44.5^{\circ}$. Therefore, over this section, the horizontal and vertical displacements are, respectively:
$$
\begin{aligned}
H & =\cos 30.5^{\circ}+\cos 31.5^{\circ}+\ldots+\cos 44.5^{\circ} \\
V & =\sin 30.5^{\circ}+\sin 31.5^{\circ}+\ldots+\sin 44.5^{\circ}
\end{aligned}
$$

However, we can also see that, letting $R$ be the circumradius of this polygon:
$$
\begin{array}{c}
H=R\left(\sin 45^{\circ}-\sin 30^{\circ}\right) \\
V=R\left[\left(1-\cos 45^{\circ}\right)-\left(1-\cos 30^{\circ}\right)\right]
\end{array}
$$

From these, we can easily compute that our desired answer is $\frac{H}{V}=(\sqrt{2}-1)(\sqrt{3}+\sqrt{2})=2-\sqrt{2}-\sqrt{3}+\sqrt{6}$.
17. $[\mathbf{1 0}]$ Compute the value of
$$
\frac{\cos 30.5^{\circ}+\cos 31.5^{\circ}+\ldots+\cos 44.5^{\circ}}{\sin 30.5^{\circ}+\sin 31.5^{\circ}+\ldots+\sin 44.5^{\circ}}
$$