696. Find the slope of the line that forms an angle of $30^{\circ}$ with the horizontal.
$\Delta$ The same proportion between the legs of a right triangle with angles of $30^{\circ}$ and $60^{\circ}$, applied in reverse, gives the answer: $k=1 / \sqrt{3}$. $\triangleleft$

There are other "convenient" angle values where the slope can be expressed by a formula (with square roots). This can be done for an angle of $72^{\circ}$ or $15^{\circ}$ (see problems 706 and 707). However, for an angle of $20^{\circ}$, there is no formula with square roots for the slope (and this is related to the fact that an angle of $60^{\circ}$ cannot be divided into three equal parts using a compass and straightedge). However, this slope can be described as the root of a cubic equation with integer coefficients (and in general, for an angle of an integer or rational number of degrees, the slope will be the root of some polynomial with integer coefficients). In practice, programs are used for approximate calculation of the slope from the angle; in a calculator, this is done using the "tg" (tangent) key.

So far, we have been talking about lines passing through the origin. Other lines are obtained

![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-206.jpg?height=277&width=297&top_left_y=795&top_left_x=1439)
from them by shifting.
![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-207.jpg?height=1706&width=392&top_left_y=347&top_left_x=338)