In a right-angled triangle, $b+c=25, \alpha=61^{\circ} 55^{\prime} 40^{\prime \prime}$. Solve the triangle and calculate the volume of the solid that is formed when the triangle rotates around its hypotenuse,
Given $b=c \cos \alpha$, therefore

$$
b+c=c(1+\cos \alpha)=25
$$

and thus

$$
c=\frac{25}{1+\cos \alpha}=\frac{25}{2 \cos ^{2} \frac{\alpha}{2}}
$$

Substituting the given values: $c=17, b=8, a=15$. The volume of the double cone formed by the rotation, where $m$ is the perpendicular from the right angle vertex to the hypotenuse:

$$
v=\frac{m^{2} \pi c}{3}=\frac{m^{2} c^{2} \pi}{3 c}=\frac{a^{2} b^{2} \pi}{3 c}=887
$$

(Petrik Sanolta, Budapest.)

The problem was also solved by: Gedliczka H., Heimlich P., Jánosy Gy., Kiss J., Martini J., Pichler S., Róth A., Schuster Gy., Szécsi I.