A right circular cone has a volume of $K=20 \mathrm{~cm}^{3}$. What is the volume of a triangular pyramid that can be inscribed in this cone, where the angles of the base triangle are:

$$
\alpha=25^{\circ} 44^{\prime} 12^{\prime \prime}, \quad \beta=82^{\circ} 12^{\prime} 40^{\prime \prime}, \quad \gamma=72^{\circ} 3^{\prime} 8^{\prime \prime} ?
$$
The height of the cone and pyramid

$$
m=\frac{3 K}{r^{2} \pi}
$$

The area of the base of the pyramid

$$
t=2 r^{2} \sin \alpha \sin \beta \sin \gamma
$$

thus, the sought volume

$$
k=\frac{2 K}{\pi} \sin \alpha \sin \beta \sin \gamma
$$

Substituting the given numerical values, we obtain

$$
k=5.21 \mathrm{~cm}^{3}
$$

(Zichy Róbert gr., Szombathely.)

Number of solutions: 44.