12.205. In the base of a right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$ lies a right triangle $A B C$, where the larger leg $A B$ is equal to $a$, and the angle opposite to it, $C$, is $\alpha$. The hypotenuse $B C$ is the diameter of the base of a cone, whose vertex lies on the edge $A_{1} B_{1}$. Find the height of the cone if $A A_{1}=a / 2$.
Solution.

Let $M$ be the vertex, $L$ the center of the base, and $M B$ and $M C$ the generators of the cone (Fig. 12.72) from the problem statement. The cone is positioned such that its height $M L$ is perpendicular to $B C$. Drop a perpendicular $M N$ from point $M$ to the plane $A B C$. Then $M N \parallel A A_{1}, M N \parallel A A_{1}, M N = A A_{2} = \frac{\alpha}{2}$, and $N L$ is the projection of $M L$ onto the plane $A B C$, with $N L \perp B C$.

$\angle B N L = \angle A C B = \alpha$.

In $\triangle B A C\left(\angle B A C = 90^{\circ}\right)$:

$$
B C = \frac{A B}{\sin \angle A C B} = \frac{a}{\sin \alpha}. \quad L B = \frac{1}{2} B C = \frac{a}{2 \sin \alpha}.
$$

In $\triangle B L N\left(\angle B L N = 90^{\circ}\right): L N = B L \operatorname{ctg} \angle B N L = \frac{a \operatorname{ctg} \alpha}{2 \sin \alpha}$.

From $\triangle M N L\left(\angle M N L = 90^{\circ}\right)$:

$$
\begin{aligned}
& M L = \sqrt{L N^{2} + M N^{2}} = \sqrt{\frac{a^{2} \operatorname{ctg}^{2} \alpha}{4 \sin ^{2} \alpha} + \frac{a^{2}}{4}} = \frac{a \sqrt{\operatorname{ctg}^{2} \alpha + \sin ^{2} \alpha}}{2 \sin \alpha}. \\
& \text{Answer: } \frac{a \sqrt{\operatorname{ctg}^{2} \alpha + \sin ^{2} \alpha}}{2 \sin \alpha}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0953.jpg?height=618&width=666&top_left_y=40&top_left_x=353)

Fig. 12.73