12.072. The base of the pyramid is a right-angled triangle with an acute angle $\alpha$. The height of the pyramid is $H$. All lateral edges form the same angle with the base plane, which is equal to $\beta$. Find the volume of the pyramid.
## Solution.

Since all the lateral edges are inclined at the same angle to the plane of the base, the base of the height of the pyramid will be the center of the circle circumscribed around the base. Since the base is a right triangle, the center of the circle circumscribed around it will be the midpoint of the hypotenuse (Fig. 12.74). Then, \( S K = H, \angle S C K = \beta, \angle A C B = 90^{\circ}, \angle C A B = \alpha \). From \(\triangle S C K\), \( C K = R = H / \operatorname{tg} \beta \), the hypotenuse \( A B = 2 R = \frac{2 H}{\operatorname{tg} \beta}, A C = A B \cos \alpha = \frac{2 H \cos \alpha}{\operatorname{tg} \beta} \). The volume of the pyramid \( V = \frac{1}{3} S_{\text {base}} \cdot H \), where \( S_{\text {base}} = \frac{1}{2} A B \cdot A C \sin \alpha = \)

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-769.jpg?height=540&width=436&top_left_y=78&top_left_x=172)

Fig. 12.74

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-769.jpg?height=542&width=402&top_left_y=61&top_left_x=749)

Fig. 12.75

\( = \frac{1}{2} \cdot \frac{2 H}{\operatorname{tg} \beta} \cdot \frac{2 H \cos \alpha}{\operatorname{tg} \beta} \sin \alpha \). Then the volume \( V = \frac{H^{2} \sin 2 \alpha}{3 \operatorname{tg}^{2} \beta} \cdot H = \frac{H^{3} \sin 2 \alpha}{3 \operatorname{tg}^{2} \beta} \).

Answer: \( \frac{H^{3} \sin 2 \alpha}{3 \operatorname{tg}^{2} \beta} \).