A person traveling in a hot air balloon notices an object on the ground to the south of them and finds that the angle of depression (the angle between the line of sight from the eye to the object and the horizontal plane through the eye) is $35^{\circ} 30^{\prime}$. The hot air balloon travels 2500 m. to the east at the same altitude above the ground, when the angle of depression to the object is $23^{\circ} 14^{\prime}$. How high is the hot air balloon flying?
Let $A$ denote the position of the object, $B$ and $C$ the positions of the barometer at the first and second locations, and $D$ a point above $A$ at the desired height. From the right triangles $ABD$, $ACD$, and $BCD$, the following relationships can be established:

\[
\begin{gathered}
B D = m \cot \beta \\
C D = m \cot \gamma \\
B C^{2} = C D^{2} - B D^{2}
\end{gathered}
\]

where

\[
\beta = 35^{\circ} 30^{\prime}, \quad \gamma = 23^{\circ} 14^{\prime} \text{ and } B C = 2500 \text{ m}
\]

From the above system of equations, it follows that:

\[
\begin{aligned}
B C^{2} & = m^{2} \left( \cot^{2} \gamma - \cot^{2} \beta \right) \\
m^{2} & = \frac{B C^{2}}{\left( \cot^{2} \gamma - \cot^{2} \beta \right)}
\end{aligned}
\]

To calculate $m$, the right-hand side of equation (4) must be transformed into a form that allows us to determine the value of $m$ using logarithmic calculations. For this purpose, I write:

\[
\cot^{2} \gamma - \cot^{2} \beta = \cot^{2} \gamma \left( 1 - \frac{\cot^{2} \beta}{\cot^{2} \gamma} \right)
\]

and

\[
\frac{\cot^{2} \beta}{\cot^{2} \gamma} = \sin^{2} \varphi
\]

which is correct because $\cot^{2} \beta < \cot^{2} \gamma$.

After these transformations, the value of the desired height is

\[
m = \frac{B C}{\cot \gamma \cos \varphi}
\]

\[
\log m = \log B C - \log \cot \gamma - \log \cos \varphi
\]

and

\[
\begin{gathered}
\log \sin \varphi = \log \cot \beta - \log \cot \gamma = \\
= 0.14673 - 0.36725 = \\
= 9.77948 - 10 \\
\varphi = 37^{\circ} 0^{\prime} 6^{\prime \prime}
\end{gathered}
\]

\[
\log m = 3.39794 - 0.36725 - 9.90234 + 10
\]

\[
\log m = 3.12835 ; \quad m = 1344 \text{ meters}
\]

(Suták Sándor, 8th grade student, Nyíregyháza).

The problem was also solved by: Visnya Aladár