Two towers stand on a slope. The top of the $56 \mathrm{~m}$ high tower sees the $42 \mathrm{~m}$ high tower at an angle of $\varphi=15^{\circ} 58^{\prime}$. (The height of the towers is understood from the base point on the slope.) The base point of the taller tower on the slope is $10 \mathrm{~m}$ higher than the base point of the shorter tower. What is the angle of inclination of the slope?
Let's create an outline for the problem. The base points of the towers are $A$ and $B$, and their tops are $C$ and $D$, respectively. The angle of inclination of the slope to the horizontal is $\alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8659b9994815c870cf0dg-1.jpg?height=634&width=458&top_left_y=252&top_left_x=823)

The two towers are: $A C=42 \mathrm{~m}, B D=56 \mathrm{~m}$. Let the angle of view from the top of the taller tower to the segment $A B$ be denoted by $\varepsilon$, and the common length of the segments $C F$ and $A E$ projected onto the line $B D$ be denoted by $u$. From the right triangles $C F D$, $A E D$, and $A E B$:

$$
\begin{gathered}
\operatorname{tg}(\varphi+\varepsilon)=\frac{u}{66-42}=\frac{u}{24} \\
\operatorname{tg} \varepsilon=\frac{u}{66} \\
\operatorname{ctg} \alpha=\frac{u}{10}
\end{gathered}
$$

Transform the (1) equation and use (2):

$$
\operatorname{tg}(\varphi+\varepsilon)=\frac{\operatorname{tg} \varphi+\operatorname{tg} \varepsilon}{1-\operatorname{tg} \varphi \operatorname{tg} \varepsilon}=\frac{\operatorname{tg} \varphi+\frac{u}{66}}{1-\frac{u}{66} \operatorname{tg} \varphi}=\frac{66 \operatorname{tg} \varphi+u}{66-u \operatorname{tg} \varphi}=\frac{u}{24}
$$

Multiply the resulting equality by the product of the denominators and rearrange:

$$
u^{2} \operatorname{tg} \varphi-42 u+1584 \operatorname{tg} \varphi=0
$$

To have $\varphi$'s trigonometric function appear in only one term, divide by $\operatorname{tg} \varphi$ and substitute $u$ with the value $u=10 \operatorname{ctg} \alpha$ from equation (3):

$$
100 \operatorname{ctg}^{2} \alpha-420 \operatorname{ctg} \varphi \operatorname{ctg} \alpha+1584=0
$$

Dividing by 100:

$$
\operatorname{ctg}^{2} \alpha-4.2 \operatorname{ctg} \varphi \operatorname{ctg} \alpha+15.84=0
$$

From this,

$$
\operatorname{ctg} \alpha=\frac{4.2 \operatorname{ctg} \varphi \pm \sqrt{17.64 \operatorname{ctg}^{2} \varphi-4 \cdot 15.84}}{2}=2.1 \operatorname{ctg} \varphi \pm \sqrt{4.41 \operatorname{ctg}^{2} \varphi-15.84}
$$

Substitute the value $\operatorname{ctg} \varphi \approx 3.495$ read from the logarithm table. Since this is a value rounded to three decimal places, we will perform the remaining calculations with three decimal place precision, and thus we get:

$$
\operatorname{ctg} \alpha \approx 7.339 \pm 6.167
$$

From this,

$$
\begin{array}{ll}
\operatorname{ctg} \alpha_{1} \approx 13.506, & \alpha_{1} \approx 4^{\circ} 14^{\prime} \\
\operatorname{ctg} \alpha_{2} \approx 1.173, & \alpha_{2} \approx 40^{\circ} 27
\end{array}
$$

Thus, the angle of inclination of the slope is either $4^{\circ} 14^{\prime}$ or $40^{\circ} 27^{\prime}$. - The latter value is quite unrealistic from a practical standpoint, as a slope of $40^{\circ}$ is not typically used for building towers; this is why the problem mentions that it is a gentle slope. Therefore, the only valid solution is the angle of inclination of $4^{\circ} 14^{\prime}$.