12.108. 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 $m \leq n$.
Solution.

By the condition $\frac{V_{1}}{V_{2}}=\frac{m}{n}$. Volume

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-794.jpg?height=418&width=426&top_left_y=863&top_left_x=743)

Fig. 12.108

$V_{1}=S \cdot H=\frac{1}{2} K P \cdot K O \cdot A D$ (Fig. 12.108). From $\triangle O K P\left(\angle O K F=90^{\circ}\right):$ $O P=\frac{K P}{\cos \alpha}, O K=O P \sin \alpha=K P \operatorname{tg} \alpha . A D=K P$. Hence, $V_{1}=\frac{1}{2} K P^{3} \operatorname{tg} \alpha$. Volume of the cube $V=A B^{3}=K P^{3}$. Then $V_{2}=V-V_{1}=K P^{3}-\frac{1}{2} K P^{3} \operatorname{tg} \alpha=$ $=K P^{3}\left(1-\frac{1}{2} \operatorname{tg} \alpha\right)$. Substitute into the ratio:
$\frac{m}{n}=\frac{V_{1}}{V_{2}}=\frac{\frac{1}{2} K P^{3} \operatorname{tg} \alpha}{\left(1-\frac{1}{2} \operatorname{tg} \alpha\right) K P^{3}}=\frac{\operatorname{tg} \alpha}{2-\operatorname{tg} \alpha}=\frac{\frac{1}{2} K P^{3} \operatorname{tg} \alpha}{\left(1-\frac{1}{2} \operatorname{tg} \alpha\right) K P^{3}}=\frac{\operatorname{tg} \alpha}{2-\operatorname{tg} \alpha}$. From this, $2 m-m \operatorname{tg} \alpha=n \operatorname{tg} \alpha ; \operatorname{tg} \alpha=\frac{2 m}{m+n} \Rightarrow \alpha=\operatorname{arctg}\left(\frac{2 m}{m+n}\right)$.

Answer: $\quad \operatorname{arctg}\left(\frac{2 m}{m+n}\right)$.