12.122. Find the volume of a cone if a chord of length $a$ in its base subtends an arc of $\alpha$ radians, and the height of the cone makes an angle $\beta$ with the slant height.
Solution.

Let $SAK$ be a cone, $SO$ its height, $BK$ a chord, $BK=a$, $\cup B n K=\alpha, \angle O S K=\beta$ (Fig. 12.122). Since $\cup B n K=\alpha$, then $\angle B O K=\alpha$. Let $R$ be the radius of the base of the cone; then in $\triangle O B K$ we have $a^{2}=2 R^{2}-2 R^{2} \cos \alpha=4 R^{2} \sin ^{2} \frac{\alpha}{2}$, i.e., $a=2 R \sin \frac{\alpha}{2}$, from which $R=\frac{a}{2 \sin (\alpha / 2)}$. From $\triangle S O K$ we find $S O=O K \operatorname{ctg} \beta=\frac{a \operatorname{ctg} \beta}{2 \sin (\alpha / 2)}$. Therefore,

$$
V=\frac{1}{3} \pi R^{2} \cdot S O=\frac{1}{3} \pi \frac{a^{2}}{4 \sin ^{2} \frac{\alpha}{2}} \cdot \frac{a \operatorname{ctg} \beta}{2 \sin \frac{\alpha}{2}}=\frac{\pi a^{3} \operatorname{ctg} \beta}{24 \sin ^{3} \frac{\alpha}{2}}
$$

$$
\text { Answer: } \frac{\pi a^{3} \operatorname{ctg} \beta}{24 \sin ^{3} \frac{\alpha}{2}}
$$