58. In the tetrahedron $ABCD$, it is given that $A \widehat{BC} = \widehat{BAD} = 90^{\circ}$, $|AB| = a$, $|DC| = b$, and the angle between the edges $AD$ and $BC$ is $\alpha$. Find the radius of the circumscribed sphere.
58. Let's take $C_{1}$ such that $A B C C_{1}$ is a rectangle (Fig. 2). $D_{1}$ is the midpoint of $A C_{1}$, $O_{1}$ and $O_{2}$ are the centers of the circumcircles of triangles $A C_{1} D$ and $A B C$, respectively, and $O$ is the center of the sphere circumscribed around $A B C D$. Clearly, $O_{2}$ is the midpoint of $A C$, and $A B$ and $C_{1} C$ are perpendicular to $A D$ and $A C_{1}$, respectively. Therefore, the planes $A D C_{1}$ and $A B C C_{1}$ are perpendicular, and $O_{1} D_{1} O_{2} O$ is a rectangle. Thus, 

$$
\left|D C_{1}\right|=\sqrt{|D C|^{2}-\left|C_{1} C\right|^{2}}=\sqrt{b^{2}-a^{2}},
$$

the radius of the circumcircle of $\triangle D C_{1} A$ will be

$$
R_{1}=\frac{\left|D C_{1}\right|}{2 \sin \widehat{D A C_{1}}}=\frac{\sqrt{b^{2}-a^{2}}}{2 \sin \alpha}
$$

The radius of the sphere $R=|O A|$ can be found from the right triangle $A O_{1} O$ (this triangle is not shown in the figure):

$$
\begin{aligned}
R=\sqrt{\left|A O_{1}\right|^{2}+\left|O_{1} O\right|^{2}}=\frac{1}{2} \sqrt{\frac{b^{2}-a^{2}}{\sin ^{2} \alpha}+a^{2}} & = \\
& =\frac{1}{2 \sin \alpha} \sqrt{b^{2}-a^{2} \cos ^{2} \alpha}
\end{aligned}
$$