12.120. The lateral edge of a regular quadrilateral pyramid is equal to $m$ and is inclined to the base plane at an angle $\alpha$. Find the volume of the pyramid.
Solution.

The volume of the pyramid $V=\frac{1}{3} S_{\text {base }} H$ (Fig. 12.120). From $\triangle S O A$ we find: $S O=H=m \sin \alpha ; A O=m \cos \alpha=\frac{A C}{2}$. Hence, $A C=2 m \cos \alpha$. The area of the base $S_{\text {base }}=\frac{A C^{2}}{2}=\frac{4 m^{2} \cos ^{2} \alpha}{2}=2 m^{2} \cos ^{2} \alpha$. Then $V=\frac{1}{3} 2 m^{2} \cos ^{2} \alpha \cdot m \sin \alpha=\frac{1}{3} m^{3} \cos \alpha \sin 2 \alpha$.

Answer: $\frac{1}{3} m^{3} \cos \alpha \sin 2 \alpha$.