$[\quad \underline{\text { Cone }}$

The vertices of two cones with a common base of radius $R$ and heights equal to $H$ and $h$ are located on opposite sides of the base. Find the angle and the distance between two generatrices of these cones, given that their ends on the circumference of the base limit a quarter of the circumference.
Let $A$ and $B$ be the vertices of the given cones, $A K$ and $B M$ be the indicated generators, and $O$ be the center of the base. According to the problem, $A O=H, B O=h, \angle K O M=90^{\circ}$ (Fig.1). Drop a perpendicular $O C$ from the center of the base to the line $A K$. Since $O K \perp O M$, by the theorem of three perpendiculars, $A K \perp O M$. Therefore, the line $A K$ is perpendicular to the plane $O C M$. When orthogonally projecting onto this plane, the line $A K$ will become the point $C$, and the point $B$ will become some point $B^{\prime}$ (Fig.2). The planes $O C M$ and $A K B$ are perpendicular because the plane $O C M$ passes through the perpendicular $O M$ to the plane $A K B$. Therefore, the perpendicular $B B^{\prime}$ to the plane $O C M$ lies in the plane $A K B$, and the point $B^{\prime}$ lies on the line $O C$ of intersection of these planes. Consider the plane $A K B$. From the right triangle $A O K$, we find that

$$
\begin{gathered}
A K=\sqrt{A O^{2}+O K^{2}}=\sqrt{H^{2}+R^{2}}, O C=\frac{A O \cdot O K}{A K}=\frac{H \cdot R}{\sqrt{H^{2}+R^{2}}} \\
A C=\frac{A O^{2}}{A K}=\frac{H^{2}}{\sqrt{H^{2}+R^{2}}}
\end{gathered}
$$

Since $B B^{\prime} \perp O C$ and $A K \perp O C$, then $B B^{\prime} \parallel A K$. Triangles $O B^{\prime} B$ and $O C A$ are similar with a ratio of $\frac{h}{H}$, therefore

$$
O B^{\prime}=\frac{h}{H} \cdot O C=\frac{h R}{\sqrt{H^{2}+R^{2}}}, B B^{\prime}=\frac{h}{H} \cdot A C=\frac{h H}{\sqrt{H^{2}+R^{2}}}.
$$

Thus,

$$
B^{\prime} C=O C+O B^{\prime}=\frac{H R}{\sqrt{H^{2}+R^{2}}}+\frac{h R}{\sqrt{H^{2}+R^{2}}}=\frac{R(H+h)}{\sqrt{H^{2}+R^{2}}}
$$

Let $\alpha$ be the angle between the lines $A K$ and $B M$. Since $B^{\prime} M$ is the orthogonal projection of the line $B M$ onto the plane $O C M$, then $\alpha=90^{\circ}-\angle B M B^{\prime}$. Therefore,

$$
\cos \alpha=\sin \angle B M B^{\prime}=\frac{B B^{\prime}}{B M}=\frac{\frac{h H}{\sqrt{H^{2}+R^{2}}}}{\sqrt{h^{2}+R^{2}}}=\frac{h H}{\sqrt{(H^{2}+R^{2})(h^{2}+R^{2})}}.
$$

From the right triangle $B M B^{\prime}$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-06.jpg?height=91&width=983&top_left_y=492&top_left_x=538)

Since the point $B^{\prime}$ lies on the line $O C$, then $O M \perp B^{\prime} C$. Therefore, $O M$ is the height of the triangle $B^{\prime} C M$. The distance between the lines $A K$ and $B M$ is equal to the length of the perpendicular $C P$ dropped from the point $C$ to the orthogonal projection $B^{\prime} M$ of the line $B M$ onto the plane $O C M$, which is perpendicular to the line $A K$. Since $C P$ is also the height of the triangle $B^{\prime} C M$, then

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-06.jpg?height=146&width=650&top_left_y=796&top_left_x=704)

## Answer

$\frac{R(H+h)}{\sqrt{H^{2}+h^{2}+R^{2}}} ; \arccos \frac{h H}{\sqrt{(H^{2}+R^{2})(h^{2}+R^{2})}}$.