The base of a straight prism is an equilateral triangle $A B C$. We draw the plane $A B D$ through the edge $A B$ such that it forms an angle of $S=35^{\circ} 18^{\prime}$ with the base, and the volume of the pyramid cut off from the prism is $V=15 \mathrm{~cm}^{3}$. What is the area of the section $A B D$?
The volume of the pyramid $ABCD$ is:

$$
V=\frac{a^{2}}{4} \sqrt{3} \frac{DC}{3}
$$

But

$$
DC=CE \cdot \tan \alpha
$$

$$
CE=a \sin 60^{\circ}=\frac{a}{2} \sqrt{3}
$$

so

$$
DC=\frac{a}{2} \sqrt{3} \tan \alpha
$$

Substituting (4) into (1), we get:

$$
V=\frac{a^{2}}{4} \sqrt{3} \cdot \frac{a}{6} \sqrt{3} \tan \alpha=\frac{a^{3}}{8} \tan \alpha
$$

from which

$$
a=2 \sqrt[3]{\frac{V}{\tan \alpha}}=2 \sqrt[3]{V \cot \alpha}
$$

The area of triangle $ABD$ is:

$$
t=\frac{a}{2} \cdot DE
$$

But

$$
DE=\frac{CE}{\cos \alpha}=\frac{a}{2 \cos \alpha} \sqrt{3}
$$

so the area of the triangle is:

$$
t=\frac{a^{2} \sqrt{3}}{4 \cos \alpha}
$$

Substituting (6) into (9), we get:

$$
t=\frac{\sqrt{3} \sqrt[3]{V^{2} \cot ^{2} \alpha}}{\cos \alpha}=\frac{\sqrt{3}}{\sin \alpha} \sqrt[3]{V^{2} \tan \alpha}
$$

Substituting the given values, we obtain:

$$
t=16.25 \text{ cm}^{2}
$$

(József Suschnik, real.VIII. o.t., Kecskemét.)

The problem was also solved by: Bernát Friedmann, S.A.-Ujhely; Nándor Kántor, Budapest; Aladár Visnya, Pécs.