12.201. The area of a lateral face of a regular dodecagonal pyramid is $S$. The planar angle at the vertex is $\alpha$. Find the volume of the pyramid.
Solution.

Let $C$ be the vertex of a regular dodecagonal pyramid (Fig. 12.68), $AB$ be a side of its base, $\angle ACB = \alpha$, $S_{\triangle ACB} = S$, $CO$ be the height of the pyramid, $CA = d$, $\angle CAO = x$.

Then $\frac{1}{2} d^{2} \sin \alpha = S$, $d = \sqrt{\frac{2 S}{\sin \alpha}}$,

$\angle CAB = 90^{\circ} - \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0948.jpg?height=735&width=430&top_left_y=319&top_left_x=835)

Fig. 12.68

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

$AO = d \cos x$, $CO = d \sin x$.

$S_{\text{base}} = 12 S_{\triangle AOB} = 12 \cdot \frac{1}{2} AO^{2} \sin \angle AOB = 6 d^{2} \cos^{2} x \sin 30^{\circ} = 3 d^{2} \cos^{2} x$.

The volume of the pyramid $V = \frac{1}{3} S_{\text{base}} \cdot CO = d^{3} \cos^{2} x \sin x$.

$\angle CAO$ is the angle of inclination of the lateral edge of the pyramid to the base plane, $\angle OAB$ is the angle between the projection of the lateral edge on the base plane and the side of the base, $\angle CAB$ is the angle between the lateral edge and the side of the base.

Then $\cos \angle CAB = \cos \angle CAO \cdot \cos \angle OAB$;

$$
\begin{aligned}
& \cos x = \frac{\cos \angle CAB}{\cos \angle OAB} = \frac{\cos \left(90^{\circ} - \frac{\alpha}{2}\right)}{\cos 75^{\circ}} = \frac{\sin \frac{\alpha}{2}}{\sin 15^{\circ}} \Rightarrow V = d^{3} \cos^{2} x \sqrt{1 - \cos^{2} x} = \\
& = \frac{2 S}{\sin \alpha} \sqrt{\frac{2 S}{\sin \alpha}} \cdot \frac{\sin^{2} \frac{\alpha}{2}}{\sin^{2} 15^{\circ}} \sqrt{1 - \frac{\sin^{2} \frac{\alpha}{2}}{\sin^{2} 15^{\circ}}} = \frac{2 S \sin^{2} \frac{\alpha}{2}}{\sin \alpha \sin^{2} 15^{\circ}} \cdot \sqrt{\frac{2 S}{\sin \alpha}} \times
\end{aligned}
$$

$$
\begin{aligned}
& \times \sqrt{\frac{\sin^{2} 15^{\circ} - \sin^{2} \frac{\alpha}{2}}{\sin^{2} 15^{\circ}}} = \frac{S \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} \sin^{3} 15^{\circ}} \cdot \sqrt{\frac{2 S}{\sin \alpha} \left(\frac{1 - \cos 30^{\circ}}{2} - \frac{1 - \cos \alpha}{2}\right)} = \\
& = \frac{S \tan \frac{\alpha}{2}}{\sin^{3} 15^{\circ}} \sqrt{\frac{2 S}{\sin \alpha} \cdot \frac{\cos \alpha - \cos 30^{\circ}}{2}} = \\
& = \frac{S \tan \frac{\alpha}{2}}{\sin^{3} 15^{\circ}} \sqrt{\frac{2 S \sin \left(15^{\circ} + \frac{\alpha}{2}\right) \sin \left(15^{\circ} - \frac{\alpha}{2}\right)}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} \cdot \frac{\sin \frac{\alpha}{2}}{2}} = \\
& = \frac{S \tan \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \sin^{3} 15^{\circ}} \sqrt{S \tan \frac{\alpha}{2} \sin \left(15^{\circ} + \frac{\alpha}{2}\right) \sin \left(15^{\circ} - \frac{\alpha}{2}\right)} \\
& \text{Result: } \frac{S}{\cos \frac{\alpha}{2} \sin^{3} 15^{\circ}} \sqrt{S \tan \frac{\alpha}{2} \sin \left(15^{\circ} + \frac{\alpha}{2}\right) \sin \left(15^{\circ} - \frac{\alpha}{2}\right)} .
\end{aligned}
$$