3.27. Lines $A C$ and $B D$, the angle between which is $\alpha\left(\alpha<90^{\circ}\right)$, touch a sphere of radius $R$ at diametrically opposite points $A$ and $B$. Line $C D$ also touches the sphere, and the angle between $A B$ and $C D$ is $\varphi\left(\varphi<90^{\circ}\right)$. Find the volume of the tetrahedron $A B C D$.
3.27. Let $A C=x, B D=y ; D_{1}$ be the projection of $D$ onto the plane tangent to the sphere at point $A$. In triangle $C A D_{1}$, angle $A$ is equal to $\alpha$ or $180^{\circ}-\alpha$, so $x^{2}+y^{2} \mp 2 x y \cos \alpha=C D_{1}^{2}=$ $=4 R^{2} \operatorname{tg}^{2} \varphi$. It is also clear that $x+y=C D=2 R / \cos \varphi$. Therefore, either $x y=R^{2} / \cos ^{2}(\alpha / 2)$, or $x y=R^{2} / \sin ^{2}(\alpha / 2)$. Considering that $(x+y)^{2} \geqslant 4 x y$, we get that the first solution is possible when $\varphi \geqslant \alpha / 2$, and the second one when $\varphi \geqslant(\pi-\alpha) / 2$. Since the volume $V$ of the tetrahedron $A B C D$ is $x y R \sin \alpha / 3$, the final answer is as follows: if $\alpha \leqslant 2 \varphi<\pi-\alpha$, then $V=2 R^{3} \operatorname{tg}(\alpha / 2) / 3$, and if $\pi-\alpha \leqslant 2 \varphi<\pi$, then $V$ can have values $2 R^{3} \operatorname{tg}(\alpha / 2) / 3$ and $2 R^{3} \operatorname{ctg}(\alpha / 2) / 3$.