The base plane, axis, and apex of two circular cones are common, the ratio of their aperture angles is $1: 2$. The two generators of the larger aperture angle cone form an angle $\alpha$. Project these generators perpendicularly onto the base plane, the angle of the projections is $\beta$. Determine the aperture angle of each cone, as well as the angle formed by the generators cut by the projection planes from the other cone, given $\alpha$ and $\beta$.
Let the apex of the cone be $C$, the center of the base circle be $O$, the generatrix of the larger opening angle cone be $C P_{2}$ and $C Q_{2}$, and the points on the base circle of the smaller opening angle cone corresponding to the radii $O P_{2}$ and $O Q_{2}$ be $P_{1}$ and $Q_{1}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_3d91e51c8b64f7f4eb07g-1.jpg?height=653&width=984&top_left_y=264&top_left_x=560)

Then, on the one hand, we need to determine the half-opening angles $\angle P_{2} C O = \varphi$ and $\angle P_{1} C O = \varphi / 2$, and on the other hand, the angle between the generatrices $C P_{1}$ and $C Q_{1}$, $\angle P_{1} C Q_{1} = x$, from the given $\angle P_{2} C Q_{2} = \alpha$ and $\angle P_{2} C Q_{2} = \angle P_{1} C Q_{1} = \beta$.

We use the fact that $\triangle C P_{2} O$ is a right triangle, $\triangle C P_{2} Q_{2}$ and $\triangle O P_{2} Q_{2}$, as well as $\triangle C P_{1} Q_{1}$ and $\triangle O P_{1} Q_{1}$, are isosceles triangles. For the larger opening angle,

$$
\sin \varphi = \frac{O P_{2}}{C P_{2}} = \frac{P_{2} Q_{2} : (2 \sin \beta / 2)}{P_{2} Q_{2} : (2 \sin \alpha / 2)} = \frac{\sin \alpha / 2}{\sin \beta / 2}
$$

and with this, given the numerically specified angles $\alpha$ and $\beta$, the smaller opening angle $\varphi / 2$ can also be considered determined.

Our result can be transferred to the inner cone by substituting $\varphi / 2$ for $\varphi$ and $x$ for $\alpha$:

$$
\sin \frac{\varphi}{2} = \frac{\sin x / 2}{\sin \beta / 2}
$$

and from this, for the sought angle between the new generatrices,

$$
\sin \frac{x}{2} = \sin \frac{\beta}{2} \sin \frac{\varphi}{2}
$$

Since in a real cone the generatrix $(C P_{2}, C P_{1})$ is larger than the radius of the base circle, in the triangles $\triangle C P_{i} Q_{i}$ and $\triangle O P_{i} Q_{i} (i=1,2)$, the angle at $C$ is smaller than the angle at $O$, so $\alpha < \beta$, and $x < \beta$. Thus, if the data satisfy $0^{\circ} < \alpha < \beta < 180^{\circ}$, then the right-hand sides of (1) and (2) are numbers between 0 and 1, and our problem is uniquely solvable.

Remark. It is more elegant to express our unknowns in terms of a suitable trigonometric function of the (original) data. For this, based on (1),

$$
\cos \varphi = \sqrt{1 - \sin^2 \varphi} = \sqrt{1 - \frac{\sin^2 \alpha / 2}{\sin^2 \beta / 2}} = \sqrt{\frac{\cos \alpha - \cos \beta}{1 - \cos \beta}}
$$

and thus the smaller opening angle directly,

$$
\sin \frac{\varphi}{2} = \sqrt{\frac{1}{2} - \frac{1}{2} \sqrt{\frac{\cos \alpha - \cos \beta}{1 - \cos \beta}}}
$$

from which, based on (2),

$$
\cos x = \frac{1}{2}(1 + \cos \beta + \sqrt{(1 - \cos \beta)(\cos \alpha - \cos \beta)})
$$

The smaller opening angle can also be expressed as

$$
\tan \frac{\varphi}{2} = \frac{\sqrt{1 - \cos \beta} - \sqrt{\cos \alpha - \cos \beta}}{\sqrt{1 - \cos \alpha}}
$$

The proof of these is left to the interested reader.