Example 2. From the vertex $B$ of an equilateral triangle $A B C$, a perpendicular $B K$ is erected to the plane $A B C$, such that $|B K|=|A B|$. We will find the tangent of the acute angle between the lines $A K$ and $B C$.






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Solution. Let's construct the image of the given figure and find its parametric number (Fig. 4).

Considering an arbitrary triangle $ABC$ as the image of an equilateral triangle,

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-014.jpg?height=605&width=671&top_left_y=246&top_left_x=1024)

Fig. 4, we spend two parameters. Considering the segment $BK$ as the image of the perpendicular to the plane $ABC$, we also spend two parameters, and finally, considering that $|BK|:|AB|=1:1$, we spend one more parameter; thus, $p=5$. Therefore, the given image is metrically defined, and any further metric constructions must not be arbitrary.

Let's perform the following additional constructions: in the plane $ABC$, draw a line $l \parallel (BC)$ through point $A$. Then the angle between the lines $AK$ and $BC$ is equal to the angle between the lines $AK$ and $l$. To find the angle between the lines $AK$ and $l$, it is obviously advisable to include it in some right triangle. The line $DK \perp l$ can be constructed as an oblique line, the projection of which is perpendicular to the line $l$. To construct this projection - the line $BD$ - we can use the fact that the triangle $ABC$ is equilateral and, therefore, its median $AM$ is also a perpendicular to the segment $BC$. Thus, by constructing the median $AM$ of the triangle $ABC$, and then $(BD) \parallel (AM)$ and $[DK]$, we will obtain the right triangle $ADK$. The ratio $|DK|:|AD|=\operatorname{tg} \widehat{DAK}$ will be the desired one. Setting $|AB|=a$, we find:

$$
|BK|=a,|BD|=\frac{a \sqrt{3}}{2},|DK|=\frac{a \sqrt{7}}{2},|AD|=|BM|=\frac{a}{2} \quad \text { and, }
$$

thus, $\operatorname{tg} \widehat{DAK}=\sqrt{7}$, and therefore, the tangent of the angle between the lines $AK$ and $BC$ is $\sqrt{7}$.