A plane flying at a constant speed of $390 \mathrm{~km} /$ hour in a westerly direction at a constant altitude sees an object located to the south at a depression angle of $\alpha=34^{\circ} 28^{\prime}$. After 1 minute, the depression angle to the same stationary object is $\beta=23^{\circ} 41^{\prime}$. At what altitude is the plane flying? (Depression angle: the angle between the line of sight and the horizontal plane.
In the initial position, let the airplane be at point $R_{0}$, and its projection on the horizontal plane be $R_{0}^{\prime}$. The object in the southern direction is denoted by $T$. Let $R_{0}^{\prime} T = d_{0}$ (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_c6009a82f698d69a3c1fg-1.jpg?height=447&width=699&top_left_y=270&top_left_x=694)

Figure 1

After 1 minute, the airplane reaches point $R$, and its perpendicular projection is $R^{\prime}$, with $R^{\prime} T = d$.

According to the problem, $R_{0} R = R_{0}^{\prime} R^{\prime} = \frac{390}{60} = 6.5 \text{ km}$. Let the distance $R_{0} R_{0}^{\prime} = R R^{\prime}$ be denoted by $x$. From the right triangle $R R^{\prime} T$,

$$
d_{0} = x \cot \beta
$$

From the right triangle $R_{0} R_{0}^{\prime} T$,

$$
d_{0} = x \cot \alpha
$$

By the Pythagorean theorem in the right triangle $T R_{0}^{\prime} R^{\prime}$,

$$
d^{2} - d_{0}^{2} = 6.5^{2}
$$

Thus,

$$
x^{2} (\cot^{2} \beta - \cot^{2} \alpha) = 6.5^{2}
$$

Therefore,

$$
\begin{gathered}
x = \frac{6.5}{\sqrt{\cot^{2} 23^{\circ} 41^{\prime} - \cot^{2} 34^{\circ} 28^{\prime}}} = \frac{6.5}{\sqrt{2.280^{2} - 1.457^{2}}} = \frac{6.5}{\sqrt{3.737 \cdot 0.823}} \sim \\
\sim \frac{6.5}{\sqrt{3.075}} \sim \frac{6.5}{1.75} \sim 3.71 \text{ km} \sim 3700 \text{ meters.}
\end{gathered}
$$

Remark: Using logarithm tables, the precise result is 3706 meters, but such precision is not necessary given the nature of the problem.

Ádám Antal (Bp. VII., Széchenyi g. I. o. t.)