Task A-3.5. (8 points)

The radius of the base of a circular cone is $r$. One axial section of this cone is a scalene triangle with angles $\alpha$ and $\beta(\alpha \neq \beta)$ at the diameter of the base. Express the volume of this cone using $r, \alpha$ and $\beta$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_59925bf190fc107f19b7g-12.jpg?height=320&width=471&top_left_y=1850&top_left_x=815)

## First Method.

Consider the described axial section. With the labels on the figure, we have $p = v \operatorname{ctg} \alpha$ and $q = v \operatorname{ctg} \beta$, so it follows that $2r = p + q = v(\operatorname{ctg} \alpha + \operatorname{ctg} \beta)$, or $v = \frac{2r}{\operatorname{ctg} \alpha + \operatorname{ctg} \beta}$.

Finally,

$$
V = \frac{1}{3} \cdot r^2 \pi \cdot v = \frac{2r^3 \pi}{3(\operatorname{ctg} \alpha + \operatorname{ctg} \beta)}
$$

Second Method.

If $v$ is the height of the cone and $s$ is the length of the slant height at angle $\alpha$, from the observed axial section we conclude that $v = s \cdot \sin \alpha$. By the Law of Sines, we also have $\frac{s}{2r} = \frac{\sin \beta}{\sin (\alpha + \beta)}$.

Now we easily get $v = \frac{2r \sin \alpha \sin \beta}{\sin (\alpha + \beta)}$, and finally,

$$
V = \frac{1}{3} \cdot r^2 \pi \cdot v = \frac{2r^3 \pi \sin \alpha \sin \beta}{3 \sin (\alpha + \beta)}
$$

For 4 points: the height of the cone expressed in terms of $\alpha, \beta$, and $r$.

For 8 points: the correct result and procedure.