3.37. The base of the pyramid is an equilateral triangle with side $a$. Two lateral faces of the pyramid are perpendicular to the base plane, and the equal lateral edges form an angle $\alpha$. Find the height of a right triangular prism equal in volume to the given pyramid and having a common base.
3.37. By condition, $DABC$ is a pyramid, $ABC$ is an equilateral triangle, $(ABD) \perp (ABC)$, $(BCD) \perp (ABC)$, $\angle ADC = \alpha$, $AB = a$ (Fig. 3.39). It is required to find the height of the prism whose base is $\triangle ABC$, such that $V_{\text{pr}} = V_{\text{pir}}$. The height of the prism coincides with the edge $DB$, since the perpendicular dropped from $D$ to $(ABC)$ must belong to both $(ABD)$ and $(CBD)$. Draw $DK \perp AC$; since $\triangle ADC$ is isosceles, then $KA = KC$. From $\triangle DKC$ we find that $DK = \frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$, and from $\triangle DBK$ we find that

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-522.jpg?height=473&width=500&top_left_y=106&top_left_x=656)

$$
DB = \sqrt{DK^2 - BK^2} = \sqrt{\frac{a^2 \operatorname{ctg}^2 \frac{\alpha}{2}}{4} - \frac{3a^2}{4}} = \frac{a}{2} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - 3}
$$

Therefore,

$$
V_{\text{pir}} = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot \frac{a}{2} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - 3} = \frac{a^3}{24} \sqrt{3 \left( \operatorname{ctg}^2 \frac{\alpha}{2} - 3 \right)}
$$

Since $V_{\text{pir}} = V_{\text{pr}} = \frac{a^2 \sqrt{3}}{4} H$, we have

$$
\frac{a^2 \sqrt{3}}{4} H = \frac{a^3}{24} \sqrt{3 \left( \operatorname{ctg}^2 \frac{\alpha}{2} - 3 \right)}
$$

from which

$$
H = \frac{a}{6} \sqrt{\operatorname{ctg}^2 \frac{\alpha}{2} - \operatorname{ctg}^2 \frac{\pi}{6}} = \frac{a \sqrt{\sin \left( \frac{\pi}{6} + \frac{\alpha}{2} \right) \sin \left( \frac{\pi}{6} - \frac{\alpha}{2} \right)}}{3 \sin \frac{\alpha}{2}}
$$