A building's one side wall is a square with a side length of $20 \mathrm{~m}$. From two points on a horizontal plane passing through the lower horizontal edge, we measured the vertical angles $\beta_{1}$ and $\delta_{1}$, and $\beta_{2}$ and $\delta_{2}$ of the vertical edges. Calculate the viewing angles of the lower and upper horizontal edges from these same points.

- Which point of the latter edges is closest to each measurement point, and at what elevation angle do we see the closest point of the upper edge?
- How far apart are the two detection points?

Numerical data: $\beta_{1}=40^{\circ}, \delta_{1}=38^{\circ} ; \beta_{2}=32^{\circ}, \delta_{2}=40^{\circ}$.
Solution. I. Let the base edge of the wall be $AD = a$, the vertical edges $AB, DC$, and the observation points $E_i$ where $i=1,2-$, from which the viewing angles of $AD, BC$ are $\alpha_i$ and $\gamma_i$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_657519b7ccafce4cdd32g-1.jpg?height=702&width=727&top_left_y=256&top_left_x=688)

Since $E_i$ is in the plane of the base edge, $E_iBA$ and $E_iCD$ are right triangles, and $AB = DC = a$, therefore

$$
AE_i = a \operatorname{cotg} \beta_i, \quad DE_i = a \operatorname{cotg} \delta_i; \quad BE_i = \frac{a}{\sin \beta_i}, \quad CE_i = \frac{a}{\sin \delta_i}
$$

(For simplicity, we will omit the index $i$ for now.) From the triangles $ADE$ and $BCE$, using the cosine rule,

$$
\begin{gathered}
\cos \alpha = \frac{AE^2 + DE^2 - AD^2}{2 \cdot AE \cdot DE} = \frac{\operatorname{cotg}^2 \beta + \operatorname{cotg}^2 \delta - 1}{2 \operatorname{cotg} \beta \operatorname{cotg} \delta} \\
\cos \gamma = \frac{\frac{1}{\sin^2 \beta} + \frac{1}{\sin^2 \delta} - 1}{\frac{2}{\sin \beta \sin \delta}} = \frac{\sin^2 \beta + \sin^2 \delta - \sin^2 \beta \sin^2 \delta}{2 \sin \beta \sin \delta}
\end{gathered}
$$

II. Let the point on $AD$ closest to $E_i$ be $K_i$, and the point on $BC$ closest to $E_i$ be $L_i$. These are cut by the plane through $E$ perpendicular to the edge, and since $AD \parallel BC$, both are cut by the same vertical plane, so it is enough to determine $K$. Based on the meaning of the function and the cosine rule,

$$
\begin{gathered}
AK = AE \cdot \cos \angle EAD = \frac{AE^2 + AD^2 - ED^2}{2 AD} = \\
= \frac{\alpha}{2} \left( \operatorname{cotg}^2 \beta - \operatorname{cotg}^2 \delta + 1 \right),
\end{gathered}
$$

and from the equality of two expressions for the area $t$ of triangle $ADE$,

$$
KE = \frac{2t}{AD} = \frac{EA \cdot ED \sin \alpha}{AD} = a \operatorname{cotg} \beta \operatorname{cotg} \delta \sin \alpha
$$

Thus, for the viewing angle $\eta$ of $KL$,

$$
\operatorname{cotg} \eta = \frac{EK}{KL} = \operatorname{cotg} \beta \operatorname{cotg} \delta \sin \alpha
$$

However, (3) and (5) only answer the question of the problem if $K$ lies on the segment $AD$: $0 \leq AK \leq AD$. Otherwise, if $AK > AD$, then $D$ is the point on the segment $AD$ closest to $E$, and for the segment $BC$, $B$ or $C$ respectively, and the elevation angles are $\beta$ or $\delta$.

III. The coordinates of the observation point $E_i$ in the right-angled coordinate system, where the origin is $A$, the positive x-axis is the ray $AD$, and the y-coordinates are positive in the half-plane containing $E_i$, can be determined. Thus, the distance $E_1E_2$ can be calculated using the distance formula of coordinate geometry.

IV. With the numerical data, for the observation point $E_1$: $\alpha_1 = 47.6^\circ, \gamma_1 = 36.6^\circ, AK_1 = 7.83 \text{ m}, K_1E_1 = 22.52 \text{ m}, \eta_1 = 41.6^\circ$; for the observation point $E_2$: $\alpha_2 = 38.6^\circ, \gamma_2 = 32.0^\circ, AK_2 = 21.39 \text{ m} (>AD), K_2E_2 = 23.80 \text{ m}$, the point on the upper edge closest to $E_2$ is $C$, and the elevation angle is $40^\circ$. Finally, from the coordinates, $E_1E_2 = 13.62 \text{ m}$.

Zambó Péter (Miskolc, Földes F. g. II. o. t.)

Note. The distance $E_1E_2$ can also be calculated from the triangle $AE_1E_2$ (or $DE_1E_2$), based on the pre-calculated angles $\angle E_iAD$ (or $\angle E_iDA$).