12.455 Let $A B$ be the diameter of the lower base of the cylinder, $A_{1} B_{1}$ a chord of the upper base, parallel to $\boldsymbol{A B}$. The plane passing through the lines $A B$ and $A_{1} B_{1}$ forms an acute angle $\alpha$ with the plane of the lower base of the cylinder, and the line $A B_{1}$ forms an angle $\beta$ with the same plane. Find the height of the cylinder if the radius of the base of the cylinder is $R$ (points $\boldsymbol{A}$ and $A_{1}$ lie on the same side of the line passing through the midpoints of segments $A B$ and $A_{1} B_{1}$).
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-322.jpg?height=325&width=261&top_left_y=265&top_left_x=98)

Let $D, C, P$ be the projections on the lower base of the given cylinder of points $B_{1}, A_{1}, O_{1}$, respectively, where $O_{1}$ is the midpoint of the chord $A_{1} B_{1}$. Let $O$ be the center of the lower base of the cylinder, and $x$ be the height of the cylinder. Then $A D=x \cdot \operatorname{ctg} \beta$, $O P \perp C D$ (by the property of the radius of a circle passing through the midpoint of a chord) $\Rightarrow O O_{1} \perp A B$ (Theorem 12) $\Rightarrow \angle O_{1} O P=\alpha, O P=x \cdot \operatorname{ctg} \alpha$.

$\angle A D B=90^{\circ}$ (Theorem 6), $OP$ is the height of $\triangle A D B \Rightarrow A B \cdot O P=A D \cdot B D=2 S_{A D B} \Rightarrow 2 R x \cdot \operatorname{ctg} \alpha=x \cdot \operatorname{ctg} \beta \cdot \sqrt{4 R^{2}-x^{2} \operatorname{ctg}^{2} \beta} \Rightarrow x=2 R \operatorname{tg}^{2} \beta \cdot \sqrt{\operatorname{ctg}^{2} \beta-\operatorname{ctg}^{2} \alpha}$. From this, the answer can be written in the following form:

Answer: $x=2 R \operatorname{tg} \beta \cdot \frac{\sqrt{\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)}}{\sin \alpha \cdot \cos \beta}$