From a balloon, the depression angles of locations $A$ and $B$ are $\alpha$ and $\beta$, and the distance between the two locations, $c$, is seen at an angle $\gamma$. What is the height at which the balloon is floating and what is the distance from the locations?
Let's denote the position of the balloon by $C$ and the projection of this point by $C_{1}$, then according to the cosine theorem:

$$
\overline{A B}^{2}=c^{2}=\overline{A C}^{2}+\overline{B C}^{2}-2 A C \cdot B C \cdot \cos \gamma
$$

but from the right-angled triangles $A C_{1} C$ and $B C_{1} C$:

$$
A C=\frac{C C_{1}}{\sin \alpha} \quad \text { and } \quad B C=\frac{C C_{1}}{\sin \beta}
$$

and thus

$$
c^{2}={\overline{C C_{1}}}^{2}\left(\frac{1}{\sin ^{2} \alpha}+\frac{1}{\sin ^{2} \beta}-\frac{2 \cos \gamma}{\sin \alpha \sin \beta}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_336d204de8387d01d92eg-1.jpg?height=448&width=645&top_left_y=638&top_left_x=729)

from which the sought height

$$
C C_{1}=\frac{c \sin \alpha \sin \beta}{\sqrt{\sin ^{2} \alpha+\sin ^{2} \beta-2 \sin \alpha \sin \beta \cos \gamma}}
$$

and

$$
A C=\frac{C C_{1}}{\sin \alpha}, \quad B C=\frac{C C_{1}}{\sin \beta}
$$

(Sándor Szibelth.)

The problem was also solved by: E. Freibauer, Ö. Kornis, R. Obláth.