A $20 \mathrm{~cm}$ radius cylinder touches the plane ground along the line $e$. Perpendicular to the line $e$, we place a $50 \mathrm{~cm}$ long rod against the cylinder so that the end of the rod on the ground is $40 \mathrm{~cm}$ away from the line $e$. How high is the other end of the rod?
Solution. Let the center of the circular cross-section of the cylinder be $O$, the cylinder touches the line $e$ at point $E_{1}$, and the rod $AB$ at point $E_{2}$, with $O A E_{1} \varangle=O A E_{2} \varangle=\alpha$. In the right triangle $O A E_{1}$:

![](https://cdn.mathpix.com/cropped/2024_05_02_671105d47e2cd114ab89g-1.jpg?height=372&width=507&top_left_y=264&top_left_x=798)

$$
\operatorname{tg} \alpha=\frac{20}{40}=\frac{1}{2}
$$

Let the perpendicular projection of $B$ onto the line $A E_{1}$ be $C$. From the triangle $A B C$, the desired length of side $B C$ is:

$$
B C=50 \cdot \sin 2 \alpha
$$

First, determine $\operatorname{tg} 2 \alpha$ using the known relationship, then find the value of $\sin 2 \alpha$.

$$
\left.\begin{array}{l}
\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3} \\
\operatorname{tg} 2 \alpha=\frac{\sin 2 \alpha}{\cos 2 \alpha}=\frac{\sin 2 \alpha}{\sqrt{1-\sin ^{2} 2 \alpha}}
\end{array}\right\} \Longrightarrow \frac{\sin 2 \alpha}{\sqrt{1-\sin ^{2} 2 \alpha}}=\frac{4}{3}
$$

Square and rearrange the equation: $9 \sin ^{2} 2 \alpha=16-16 \sin ^{2} 2 \alpha$. From this, $\sin 2 \alpha=\frac{4}{5}$ (only the positive root is considered). Substitute the value of $\sin 2 \alpha$ into (1): $B C=50 \cdot \frac{4}{5}=40$. Therefore, the other end of the rod is $40 \mathrm{~cm}$ high.