12.278. The base of the pyramid is an isosceles trapezoid, where the acute angle is $\alpha$, and the area is $S$. All lateral faces form the same angle $\beta$ with the base plane. Find the volume of the pyramid.
## Solution.

Let trapezoid $ABCD$ be the base of the pyramid $EABCD$ (Fig. 12.141), $AB=CD, \angle ABC=\alpha, 0^{\circ}<\alpha<90^{\circ}, EO$ - the height of the pyramid.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1033.jpg?height=790&width=753&top_left_y=33&top_left_x=318)

Fig. 12.142

Since all lateral faces form the same angle with the base plane, point $O$ is the center of the circle inscribed in trapezoid $ABCD$.

Then $AD+BC=AB+CD=2AB$. If the height of trapezoid $ABCD$ is $AF=h$, then from $\triangle AFB\left(\angle AFB=90^{\circ}\right)$:

$$
AB=\frac{h}{\sin \alpha}
$$

The area of the base of the pyramid

$$
S=\frac{AD+BC}{2} \cdot AF=AB \cdot AF=\frac{h^2}{\sin \alpha}
$$

Then $h=\sqrt{S \sin \alpha}$.

In the base plane, draw the perpendicular $OK$ to $AB$. Then $EK \perp AB \angle EKO=\beta$, the radius of the circle inscribed in trapezoid $ABCD$, $OK=\frac{1}{2} h=\frac{1}{2} \sqrt{S \sin \alpha}$.

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

$EO=KO \operatorname{tg} \angle EKO=\frac{\operatorname{tg} \beta \sqrt{S \sin \alpha}}{2}$.

The volume of the pyramid $V=\frac{1}{3} S \cdot EO=\frac{S \operatorname{tg} \beta \sqrt{S \sin \alpha}}{6}$.

Answer: $\frac{S \operatorname{tg} \beta \sqrt{S \sin \alpha}}{6}$.