$\begin{aligned} & \text { [ Regular Pyramid ] } \\ & \text { [ Volume of Tetrahedron and Pyramid ] }]\end{aligned}$

Find the volume of a regular triangular pyramid with a lateral edge $b$ and a 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

$$
\begin{gathered}
\frac{a}{2} = BK = BD \sin \angle BDK = b \sin \frac{\phi}{2} \\
DM = \sqrt{BD^2 - BM^2} = \sqrt{b^2 - \frac{a^2}{3}} = \sqrt{b^2 - \frac{1}{3} b^2 \sin^2 \frac{\phi}{2}} = b \sqrt{1 - \frac{1}{3} \sin^2 \frac{\phi}{2}}
\end{gathered}
$$

Therefore,

$$
\begin{aligned}
& V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot b \sqrt{1 - \frac{1}{3} \sin^2 \frac{\phi}{2}} = \\
& = \frac{\sqrt{3}}{12} \cdot 4 b^2 \sin^2 \frac{\phi}{2} \cdot b \sqrt{1 - \frac{1}{3} \sin^2 \frac{\phi}{2}} = \frac{1}{3} b^3 \sin^2 \frac{\phi}{2} \sqrt{3 - 4 \sin^2 \frac{\phi}{2}}.
\end{aligned}
$$

## Answer

$\frac{1}{3} b^3 \sin^2 \frac{\phi}{2} \sqrt{3 - 4 \sin^2 \frac{\phi}{2}} = \frac{1}{3} b^3 \sin^2 \frac{\phi}{2} \sqrt{1 + \cos \phi}$.