To determine the height of a mountain peak, we conducted measurements from three points, $A$, $B$, and $C$. From point $A$, the angle of elevation to the peak is $20^{\circ}$, and the direction to the base of the mountain forms a $93^{\circ}$ angle with the direction $BA$. At point $B$, we determined that the three points lie in a horizontal plane, the segment $AB$ is $100 \mathrm{~m}$, and the segment $BC$ is $150 \mathrm{~m}$. The angle between $AB$ and $BC$ is $130^{\circ}$, and the peak is within this angle. From point $C$, the angle of elevation to the peak is $22^{\circ}$. What can we determine about the height of the peak above the plane of the observation points?
Let the mountain peak be $H$, and its projection on the horizontal plane $ABC$ be $T$. Then, from the right-angled triangles $HAT$ and $HCT$, we have $TA = HT \operatorname{ctg} 20^{\circ}$ and $TC = HT \operatorname{ctg} 22^{\circ}$. Thus, from the quadrilateral $ABCT$, we know the sides $AB$ and $BC$, the angles at $B$ and $A$—the latter being the complementary angle of the measured $93^{\circ}$ angle, i.e., $87^{\circ}$—and finally the ratio of the sides meeting at $T$: $TA : TC = \operatorname{ctg} 20^{\circ} : \operatorname{ctg} 22^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b14a2a0786a0e24cbf15g-1.jpg?height=518&width=1160&top_left_y=348&top_left_x=447)

From the triangle $ABC$, using the cosine and sine theorems, we get $AC = 227.6 \, \text{m}$ and $\angle BAC = 30.34^{\circ}$, so $\angle CAT = 56.66^{\circ}$ (we wrote out 4 significant digits from the intermediate results, as the data were considered accurate).

Thus, in the triangle $ACT$, we know the angle opposite the smaller of the known ratio sides, $TC$, since $\operatorname{ctg} 22^{\circ} < \operatorname{ctg} 20^{\circ}$. Therefore, using the sine theorem based on the ratio, we can calculate the sine of $\angle ACT$—if it turns out to be less than 1, the angle can be either acute or obtuse. Since

$$
\sin \angle ACT = \sin \angle CAT \cdot \frac{\operatorname{ctg} 20^{\circ}}{\operatorname{ctg} 22^{\circ}} = 0.9272 < 1,
$$

one possible value for $\angle ACT$ is $\angle ACT_1 = 68^{\circ}$, and the other is $\angle ACT_2 = 112^{\circ}$.

From the first, we get $\angle AT_1C = 55.34^{\circ}$, and using the sine theorem, $CT_1 = 231.2 \, \text{m}$ and $T_1H_1 = 93.4 \, \text{m}$. From the second, similarly, $\angle AT_2C = 11.34^{\circ}$, $CT_2 = 967.4 \, \text{m}$, and $T_2H_2 = 390.9 \, \text{m}$.

Thus, we obtained alternative results for the height of the mountain: either $93.4 \, \text{m}$ or $390.9 \, \text{m}$ above the plane of the observation points.

Mitrocsák Anikó (Makó, József A.g. III. o. t.)

Note. Several solutions took the angle $\angle BAT$ to be $93^{\circ}$. In this case, two results are also obtained: $136 \, \text{m}$ and $268 \, \text{m}$.