3.44. Find the volume of a cone if a chord equal to $a$ in its base subtends an arc of $\alpha^{\circ}$, and the height of the cone forms an angle $\beta$ with the slant height.
3.44. Given that $SAK$ is a cone, $SO$ is its height, $BK$ is a chord, $BK=a, \cup B n K=\alpha, \angle O S K=\beta$ (Fig. 3.46). Since $\cup B n K=\alpha^{\circ}$, then

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-526.jpg?height=542&width=417&top_left_y=1290&top_left_x=122)

Fig. 3.45

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-526.jpg?height=569&width=418&top_left_y=1290&top_left_x=687)

Fig. 3.46
$\angle B O K=\alpha^{0}$. Let $R$ be the radius of the base of the cone; then in $\triangle O B K$ we have $a=2 R \sin \frac{\alpha}{2}$, from which $R=\frac{a}{2 \sin \frac{\alpha}{2}}$. From $\triangle S O K$ we find $S O=O K \operatorname{ctg} \beta=\frac{a \operatorname{ctg} \beta}{2 \sin \frac{\alpha}{2}}$. Therefore,

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

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-527.jpg?height=305&width=492&top_left_y=755&top_left_x=46)

Fig. 3.47