12.440 The ratio of two segments enclosed between parallel planes is $\boldsymbol{k}$, and the angles that each of these segments makes with one of the planes are in the ratio $2: 3$. Find these angles and the permissible values of $\boldsymbol{k}$.
Solution. Let the given segments be equal to $a$ and $k a$, and the angles they form with one of the planes be $x$ and $\frac{2}{3} x$ respectively. Since the planes are parallel, the distance between them is $a \cdot \sin x$ or $k a \cdot \sin \frac{2}{3} x$, i.e., $k \sin \frac{2}{3} x = \sin x$.

From this, $2 k \cdot \sin \frac{x}{3} \cdot \cos \frac{x}{3} = 3 \sin \frac{x}{3} - 4 \sin ^{3} \frac{x}{3} \Rightarrow 4 \cos ^{2} \frac{x}{2} - 2 k \cdot \cos \frac{x}{3} - 1 = 0 \Rightarrow$ $\Rightarrow \cos \frac{x}{3} = \frac{k + \sqrt{k^{2} + 4}}{4}$. But $0^{\circ} < x < 90^{\circ} \Rightarrow 0 < \cos \frac{x}{3} \leq \cos 30^{\circ} \Leftrightarrow 1 > \frac{k + \sqrt{k^{2} + 4}}{4} \geq \frac{\sqrt{3}}{2} \Leftrightarrow$

$$
\Leftrightarrow\left\{\begin{array}{l}
k < \frac{3}{2} \\
k^{2} + 4 \geq 4k^{2} + 4, \quad \Leftrightarrow \frac{2 \sqrt{3}}{3} \leq k < \frac{3}{2}
\end{array}\right.
$$

Answer: $\quad 2 \arccos \frac{k + \sqrt{k^{2} + 4}}{4} ; 3 \arccos \frac{k + \sqrt{k^{2} + 4}}{4} ; \frac{2 \sqrt{3}}{3} \leq k < \frac{3}{2}$.