3.27. A plane is drawn through the side of the lower base of a cube, dividing the volume of the cube in the ratio $m: n$, counting from the lower base. Find the angle between this plane and the plane of the base, if $m \leq n$.
3.27. From the condition $m \leq n$, it follows that the plane $A D N M$ intersects the face of the cube $B B_{1} C_{1} C$ (and not $A_{1} B_{1} C_{1} D_{1}$) along the line $M N \| A D$ and, therefore, $M N$ is perpendicular to the face $D D_{1} C_{1} C$ (Fig. 3.29). Let the edge of the cube be $a$, and $\angle N D C=\alpha$. Since the angle $N D C$ is the linear angle of the desired angle between the plane of the section and the lower base of the cube, then $N C=D C \operatorname{tg} \alpha=a \operatorname{tg} \alpha$ (from $\triangle D N C$) and, therefore,
$S_{\triangle D C N}=\frac{1}{2} a^{2} \operatorname{tg} \alpha . \quad$ The figure DNCAMB is a right triangular prism with the base $D N C$. Its volume

$$
V_{1}=S_{\triangle D N C} \cdot M N=\frac{1}{2} a^{3} \operatorname{tg} \alpha
$$

Since the volume of the cube $V=a^{3}$, we find $V-V_{1}=a^{3}\left(1-\frac{1}{2} \operatorname{tg} \alpha\right)$. According to the condition, we have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-516.jpg?height=553&width=539&top_left_y=108&top_left_x=613)

Fig. 3.29

$$
\begin{gathered}
\frac{V_{1}}{V-V_{1}}=\frac{m}{n} ; \frac{m}{n}=\frac{a^{3} \operatorname{tg} \alpha}{2}: a^{3}\left(1-\frac{1}{2} \operatorname{tg} \alpha\right) ; \frac{m}{n}=\frac{\operatorname{tg} \alpha}{2-\operatorname{tg} \alpha} ; \\
\frac{n}{m}=\frac{2-\operatorname{tg} \alpha}{\operatorname{tg} \alpha} ; \frac{m+n}{m}=\frac{2}{\operatorname{tg} \alpha} ; \operatorname{tg} \alpha=\frac{2 m}{m+n}
\end{gathered}
$$

from which $\alpha=\operatorname{arctg} \frac{2 m}{m+n}$.