70. Practical task. The factory territory is flat, but gently sloping. We have a leveling instrument consisting of a horizontal telescope rotating around a vertical axis (with angles of rotation read from the scale of the horizontal circle), and a staff, which we sight to read the difference in levels, as well as the distance at which the staff is set (thanks to two horizontal threads in the telescope and a scale on the staff). How can we most easily determine the slope of the land and the direction of the slope?
70. The task is solved as follows: we install a theodolite at point $O$ of the given section (Fig. 147) and measure the slope of the section in an arbitrarily chosen direction $O C_{1}$. For this purpose, we install a staff at point $A_{1}$ in the same direction and

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-144.jpg?height=896&width=968&top_left_y=660&top_left_x=544)

Fig. 146.

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-144.jpg?height=489&width=1077&top_left_y=1663&top_left_x=478)

Fig. 147.

measure the distance $d_{1}$ and the difference in levels $h_{1}$ between points $O$ and $A_{1}$. The slope of the section is determined as the quotient $t_{1}=h_{1} / d_{1}$. Then we rotate the theodolite by $90^{\circ}$ and in the direction $O C_{2}$, perpendicular to $O C_{1}$, measure the slope of the section, expressed by the number $t_{2}=h_{2} / d_{2}$.

Taking the ray $O C_{1}$ as the $O x$ axis and $O C_{2}$ as the $O y$ axis, we find the vector with components $t_{1}, t_{2}$ on these axes; the direction of this vector is the direction of the slope of the section, and its length, i.e., the number $\sqrt{t_{1}^{2}+t_{2}^{2}}$, is equal to the slope of the section.

Here is the justification for this method. Let $p$ be the contour line passing through point $O$, $q$ be the slope line, $q^{\prime}$ be the projection of the slope line on the horizontal plane $\Pi$, $\beta$ be the angle of inclination, and $\alpha$ be the unknown angle that the direction $O C_{2}$ forms with the direction $q^{\prime}$.

We have:

$$
\operatorname{tg} \beta=\frac{h_{1}}{x_{1}}=\frac{h_{2}}{x_{2}}
$$

from which

$$
\frac{x_{1}}{x_{2}}=\frac{h_{1}}{h_{2}}
$$

Moreover, we have:

$$
\frac{x_{1}}{d_{1}}=\sin \alpha, \quad \frac{x_{2}}{d_{2}}=\cos \alpha
$$

therefore

$$
\operatorname{tg} \alpha=\frac{x_{1} d_{2}}{d_{1} x_{2}}=\frac{h_{1} d_{2}}{d_{1} h_{2}}=\frac{t_{1}}{t_{2}}
$$

thus, the method of finding the direction of the slope, as described above, is correct.

Finally, we calculate the slope of the section:

$$
\operatorname{tg} \beta=\frac{h_{2}}{x_{2}}=\frac{h_{2}}{d_{2} \cos \alpha}=\frac{h_{2}}{d_{2}} \sqrt{1+\operatorname{tg}^{2} \alpha}=\sqrt{t_{1}^{2}+t_{2}^{2}}=t
$$