Example 57. At the base of the pyramid lies an equilateral triangle. One of the lateral faces is perpendicular to the base plane, while the other two are inclined to it at angles equal to $\alpha$. We need to find the angles formed by the lateral edges with the base plane.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals $S B$ and $A C$ be the image of the given pyramid (Fig. 63). This image is complete (verify this independently, for example, by taking the plane $A B C$ as the main plane). The parametric number of the image (calculate it independently) is 4.

Given: $\quad S A B C$ - pyramid, $S$ - vertex, $|A B|=|A C|=$ $=|B C|,((S A B),(A B C))=90^{\circ},((S A C), \widehat{(A B C)})=((S B C) \widehat{(A B C)})=\alpha$. Find: $((S \widehat{A}),(A B C)),((S B),(A B C)),((S \widehat{C}),(A B C))$.
S o l u t i o n. To introduce the given and sought angles into the drawing, we will perform the necessary additional constructions. We will construct $[S D]$ - the height of $\triangle S A B$.

Then, since $(S A B) \perp(A B C)$, we have $(S D) \perp(A B C)$, and therefore $[A D]$ is the projection of the edge $S A$ onto the plane $A B C$, i.e., $\angle S A D$ is the angle formed by the edge $S A$ with the plane $A B C$. Similarly, $\angle S B D$ is the angle formed by the edge $S B$ with the plane $A B C$, and $\angle S C D$ is the angle formed by the edge $S C$ with the plane $A B C$. Note further that $[A E]$ - the median of triangle $A B C$ is also its height, so by constructing $(D F) \|(A E)$, we conclude that $(D F) \perp(B C)$, and therefore $[D F]$ is the projection of the segment $S F$ onto the plane $A B C$. Thus, $[S F] \perp(B C)$ and $\angle S F D$ is the linear angle of the dihedral angle $B C$, i.e., $\widehat{S F D}=\alpha$. Similarly, by constructing $(D K) \perp(A C)$, we conclude that $\widehat{S K D}=\alpha$. Moreover, the right triangles $S D F$ and $S D K$ have a common leg $S D$ and congruent acute angles $S F D$ and $S K D$, i.e., these triangles are congruent. Then $|D F|=|D K|$. This means that point $D$ is equidistant from the sides of angle $A C B$. Thus, point $D$ lies on the bisector of angle $A C B$ of the equilateral triangle $A B C$. But the bisector of an equilateral triangle is also its median and height. Therefore, $[C D]$ is the median of $\triangle A B C$, i.e., point $D$ is the midpoint of side $A B$, and thus $[S D]$ - the height of $\triangle S A B$ is also its median. This means that $\triangle S A B$ is isosceles, i.e., $\widehat{S B D}=\widehat{S A D}$.

We will solve the problem algebraically.

Let for brevity $\widehat{S B D}=x$ and $\widehat{S C D}=y$. Then $\widehat{S A D}=x$. Since this is a problem of type $M$, we introduce the auxiliary parameter $|A B|=a$. We proceed to the calculations. We have: $|B D|=\frac{a}{2}$. Since $[A E]$ is the median of $\triangle A B C$, and $[F D]$ is the midline of $\triangle A B E$, then

$$
|D F|=\frac{1}{2}|A E|=\frac{1}{2} \cdot \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{4}
$$

From the right triangle $S D F$ we have:

$$
|S D|=|D F| \operatorname{tg} \alpha=\frac{a \sqrt{3}}{4} \operatorname{tg} \alpha
$$

94

Thus, $\operatorname{tg} x=\frac{|S D|}{|B D|}=$ $=\frac{\sqrt{3}}{2} \operatorname{tg} \alpha$

from which $x=\operatorname{arctg}\left(\frac{\sqrt{3}}{2} \operatorname{tg} \alpha\right)$.

Next, $\operatorname{tg} y=\frac{|S D|}{|C D|}=\frac{1}{2} \operatorname{tg} \alpha$,

from which $\quad y=\operatorname{arctg}\left(\frac{1}{2} \operatorname{tg} \alpha\right)$.

Investigation.

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-096.jpg?height=497&width=671&top_left_y=248&top_left_x=1024)

Fig. 64 According to the problem, $0^{\circ}<x<y<90^{\circ}$.
But since $0^{\circ}<\alpha<90^{\circ}$, we have $0<\operatorname{tg} x<\infty$, and therefore $0<\frac{\sqrt{3}}{2} \operatorname{tg} \alpha<\frac{1}{2} \operatorname{tg} \alpha<\infty$. Then for the found values of $x$ and $y$, the system of inequalities $0^{\circ}<x<y<90^{\circ}$ is true.

Thus, $((S A) \widehat{(A B C)})=\operatorname{arctg}\left(\frac{1}{2} \operatorname{tg} \alpha\right)$,

$((S B) \widehat{,}(A B C))=((S C),(A B C))=\operatorname{arctg}\left(\frac{\sqrt{3}}{2} \operatorname{tg} \alpha\right)$.