The sides of a triangle form a geometric sequence with a common ratio $q$; what are its angles? (for example, $q=\frac{3}{2}$).
Let $\alpha$ be the angle opposite to $a$, $\beta$ be the angle opposite to $a q$, and $\gamma$ be the angle opposite to $a q^{2}$. Applying Carnot's theorem:

$$
a^{2} q^{2} + a^{2} q^{4} - 2 a^{2} q^{3} \cos \alpha = a^{2}
$$

from which

$$
\cos \alpha = \frac{q^{4} + q^{2} - 1}{2 q^{3}}
$$

Similarly,

$$
\cos \beta = \frac{1 - q^{2} + q^{4}}{2 q^{2}} \text{ and } \cos \gamma = \frac{1 + q^{2} - q^{4}}{2 q}
$$

If $q = \frac{3}{2}$, then $\cos \alpha = \frac{101}{108}$, $\cos \beta = \frac{61}{72}$, and $\cos \gamma = -\frac{29}{48}$, so $\alpha = 20^{\circ} 44^{\prime} 27^{\prime \prime}$, $\beta = 32^{\circ} 05^{\prime} 22^{\prime \prime}$, and $\gamma = 127^{\circ} 10^{\prime} 11^{\prime \prime}$.

(Viktor Wáhl, Eger.)

The problem was also solved by: Bánó L., Eckhart F., Fekete M., Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Miklóssy L., Petrik S., Pichler S., Sárközy S., Schuster Gy., Steiger J., Tóth B., Winkler J.