12.248. A right prism with a rhombus as its base is circumscribed around a sphere. The larger diagonal of the prism forms an angle $\alpha$ with the base plane. Find the acute angle of the rhombus.
Solution.

Let the rhombus $ABCD$ be the base of the given right prism $ABCD A_1 B_1 C_1 D_1$, circumscribed around a sphere (Fig. 12.114), and the radius of the sphere is $R$. Then the height of the prism $BB_1 = 2R$.

If $B_1 D$ is the major diagonal of the prism, then $\angle B_1 DB = \alpha$ and $BD$ is the major diagonal of the base $\Rightarrow \angle ABC$ is the acute angle of the rhombus we are looking for.

In $\triangle B_1 BD: BD = 2R \operatorname{ctg} \alpha$.

If a perpendicular section is drawn through the midpoint of the height of the prism, a rhombus equal to the base of the prism is obtained, in which the great circle of the given sphere is inscribed. From this, the radius of the sphere is equal to the radius of the circle inscribed in the rhombus $ABCD$.

The center of this circle is the point $O$ where the diagonals of the rhombus intersect.

$E$ is the point of tangency of this circle with the side $AB$ of the rhombus.

Then $OE \perp AB, OE = R. BO = \frac{1}{2} BD = R \operatorname{ctg} \alpha. \triangle BEO (\angle BEO = 90^\circ)$: $\sin \angle OBE = \frac{OE}{OB} = \frac{R}{R \operatorname{ctg} \alpha} = \operatorname{tg} \alpha$

$\angle OBE = \arcsin (\operatorname{tg} \alpha)$

$\angle ABC = 2 \angle OBE = 2 \arcsin (\operatorname{tg} \alpha)$.

Answer: $2 \arcsin (\operatorname{tg} \alpha)$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0998.jpg?height=648&width=767&top_left_y=76&top_left_x=289)

Fig. 12.115