12.370. The base of the pyramid is a right-angled triangle, one of whose acute angles is $\alpha$. All lateral edges are equally inclined to the plane of the base. Find the dihedral angles at the base if the height of the pyramid is equal to the hypotenuse of the triangle lying in its base.
## Solution.

Let $S O$ be the height of the pyramid $S A B C, \angle A C B=90^{\circ}, \angle B A C=\alpha$ and, since $\angle S A O=\angle S B O=\angle S C O$, then $O$ is the center of the circle circumscribed around the base of the pyramid, which is the midpoint of the hypotenuse $A B$ (Fig. 12.230), thus, the plane $A S B \perp$ plane $A B C$.

$O D \perp A C, O E \perp B C$. Then $S D \perp A C, S E \perp B C ;$

$\angle S D O=\varphi$ and $\angle S E O=\gamma-$ are the linear angles of the dihedral angles at the base of the pyramid.

Let $S O=A B=1$. Then $A O=B O=\frac{1}{2}$.

In $\triangle A D O\left(\angle A D O=90^{\circ}\right): O D=O A \sin \angle A O D=\frac{\sin \alpha}{2}$.

In $\triangle B E O\left(\angle B E O=90^{\circ}\right)$ :

$$
O E=O B \sin \angle O B E=\frac{\sin \left(90^{\circ}-\alpha\right)}{2}=\frac{\cos \alpha}{2}
$$

In $\triangle S O D\left(\angle S O D=90^{\circ}\right): \operatorname{tg} \varphi=\frac{S O}{O D}=\frac{2}{\sin \alpha} ;$

$\varphi=\operatorname{arctg} \frac{2}{\sin \alpha}$.

In $\triangle S O E\left(\angle S O E=90^{\circ}\right): \operatorname{tg} \gamma=\frac{S O}{O E}=\frac{2}{\cos \alpha}$.

$$
\gamma=\operatorname{arctg} \frac{2}{\cos \alpha}
$$

Answer: $90^{\circ} ; \operatorname{arctg} \frac{2}{\sin \alpha} ; \operatorname{arctg} \frac{2}{\cos \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1132.jpg?height=664&width=573&top_left_y=31&top_left_x=97)

Fig. 12.230

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1132.jpg?height=587&width=618&top_left_y=108&top_left_x=656)

Fig. 12.231