3.34. The base of the pyramid is a right-angled triangle, the hypotenuse of which is equal to $c$, and one of the acute angles is equal to $\alpha$. Each lateral edge forms an angle $\beta$ with the plane of the base. Find the volume of the pyramid.
3.34. Since the lateral edges of the pyramid $ABCD$ (Fig. 3.36) are equally inclined to the plane of the base, the vertex $D$ projects onto the center of the circle circumscribed around triangle $ABC$; for a right triangle, it projects onto the midpoint $E$ of the hypotenuse $AB$. The height $DE$ of triangle $ADB$ is the height of the pyramid. Further,

we have $DE = BE \operatorname{tg} \beta = \frac{c}{2} \operatorname{tg} \beta$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-520.jpg?height=507&width=516&top_left_y=106&top_left_x=640)

Fig. 3.36

$$
S_{\triangle ABC} = \frac{1}{2} AC \cdot BC = \frac{1}{2} c^2 \sin \alpha \cos \alpha = \frac{1}{4} c^2 \sin 2 \alpha
$$

Finally, we find

$$
V = \frac{1}{3} \cdot \frac{1}{4} c^2 \sin 2 \alpha \cdot \frac{c}{2} \operatorname{tg} \beta = \frac{c^3}{24} \sin 2 \alpha \operatorname{tg} \beta
$$