$\begin{aligned} & {\left[\begin{array}{l}\text { Regular pyramid }\end{array}\right]} \\ & \text { [Volume of a tetrahedron and pyramid] }\end{aligned}$

Find the volume of a regular triangular pyramid with height $h$ and the plane angle $\phi$ at the apex of the pyramid.
Let $K$ be the midpoint of side $BC$ of the base $ABC$ of a regular triangular pyramid $ABCD$, and $M$ be the center of the base $ABC$. According to the problem, $\angle BDC = \phi$. Denote $AB = BC = AC = a$. From the right triangles $BKD$ and $BMD$, we find that

$$
\begin{gathered}
BD = \frac{BK}{\sin \angle BDK} = \frac{\frac{a}{2}}{\sin \frac{\phi}{2}} = \frac{a}{2 \sin \frac{\phi}{2}}, \\
BD^2 = DM^2 + BM^2 = h^2 + \frac{1}{3}a^2.
\end{gathered}
$$

From the equation

$$
\frac{a^2}{4 \sin^2 \frac{\phi}{2}} = h^2 + \frac{1}{3}a^2
$$

we find that

$$
a^2 = \frac{12 h^2 \cos^2 \frac{\phi}{2}}{3 - 4 \sin^2 \frac{\phi}{2}}.
$$

Therefore,

$$
V_{ABCD} = \frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^2 \cdot h = \frac{\sqrt{3}}{12} a^2 h = \frac{\sqrt{3}}{12} \cdot \frac{12 h^2 \cos^2 \frac{\phi}{2}}{3 - 4 \sin^2 \frac{\phi}{2}} = \frac{h^2 \sqrt{3} \cos^2 \frac{\phi}{2}}{3 - 4 \sin^2 \frac{\phi}{2}}.
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_35476b28e4766e84fc81g-29.jpg?height=91&width=349&top_left_y=2490&top_left_x=0)