A hiker sees a mountain peak from a point on a hillside at an elevation angle of $\alpha=23^{\circ} 20^{\prime}$ above the horizontal and sees the reflection of the same mountain peak in the mirror of a lake below at a depression angle of $\beta=49^{\circ} 30^{\prime}$ below the horizontal. How high is the hiker above sea level if it can be determined from the map that the mountain peak is $1827 \mathrm{~m}$ high and the surface of the lake is $608 \mathrm{~m}$ above sea level? How far is the hiker (in a straight line) from the mountain peak?
The figure shows the labeling, $s$ is the sea level, $n$ is the lake's mirror, $T$ is the tourist's position, $C$ is the peak of the mountain, $C^{\prime}$ is the reflection of the mountain peak. (Thus $N C = N C^{\prime}$).

$$
N C = S C - S N = 1827 - 608 = 1219
$$

and thus

$$
C C^{\prime} = 2 \cdot 1219 = 2438
$$

If we denote the sought $T C$ distance by $x$ and the height of the tourist above sea level by $T S^{*} = y$, then

$$
\begin{aligned}
& C T^{*} = x \sin \alpha \\
& T T^{*} = x \cos \alpha \\
& T^{*} C^{\prime} = T T^{*} \operatorname{tg} \beta = x \cos \alpha \cdot \operatorname{tg} \beta
\end{aligned}
$$

Accordingly,

$$
C T^{*} + T^{*} C^{\prime} = x \sin \alpha + \cos \alpha \operatorname{tg} \beta = C C^{\prime} = 2438
$$

from which

$$
x = \frac{2438}{\sin 23^{\circ} 20^{\prime} + \cos 23^{\circ} 20^{\prime} \cdot \operatorname{tg} 49^{\circ} 30^{\prime}} \approx 1657 \text{ m}
$$

and thus

$$
C T^{*} = x \sin \alpha = 1657 \sin 23^{\circ} 20^{\prime} = 656 \text{ m},
$$

so

$$
y = S C - C T^{*} = 1827 - 656 = 1171 \text{ m}
$$

Kántor Sándor (Debrecen, Ref. g. IV. o. t.)