Suppose that a narrow beam flash-light is spun around its center which is located a unit distance from the $x$-axis, in such a way that the angle between the beam and the line through the center, parallel to $x$-axis varies uniformly on $[0,1]$ . Consider the random point $X$ at which the beam intersects the $x$-axis when the flash-light has stopped spinning.
Then how to find the distribution of $X$ ?
I will set flashlight to be at $(0,1)$ relative to the origin for simplicity. You can just translate it later. The angle it makes with the horizontal is $\Theta$.
We can find using trigonometry that $$\tan \Theta = \frac{1}{X}$$ So $$X = \frac{1}{\tan \Theta}$$ Hence, $$P\left(X \leq x\right)=P\left(\frac{1}{\tan \Theta} \leq x \right)$$ Since both $X$ and $\tan \Theta$ are positive and $\tan \Theta$ is increasing for $0 \leq \Theta \leq 1$, $$P\left(X \leq x\right)=P\left(\Theta \geq \tan^{-1}\frac{1}{x} \right)$$ Using the CDF for unifrom distribution, we have $$P\left(X \leq x\right)=1-\tan^{-1}\frac{1}{x}$$ Hence, $$f_X(x)= \begin{cases} \frac{1}{x^2+1}, & \text{for } x \ge\cot 1 \text {,} \\ 0, & \text{otherwise.} \\ \end{cases}$$
Update: The extended case where $0 \leq \Theta \leq \pi$
For $0 \leq \Theta \leq \frac{\pi}{2}$, both $X$ and $\tan \Theta$ are positive and $\tan \Theta$ is increasing. Thus, $$P\left(X \leq x\right)=P\left(\Theta \geq \tan^{-1}\frac{1}{x} \right)$$
For $0 \leq \Theta \leq \frac{\pi}{2}$, both $X$ and $\tan \Theta$ are negative and $\tan \Theta$ is increasing. Thus, $$P\left(X \leq x\right)=P\left(\Theta \geq \tan^{-1}\frac{1}{x} \right)$$ which is exactly the same as the other subrange.
Using the CDF for unifrom distribution, we have $$P\left(X \leq x\right)=\frac{1}{\pi}\left(1-\tan^{-1}\frac{1}{x}\right)$$
Hence, $$f_X(x)=\frac{1}{\pi (x^2+1)}$$ for $x \in \mathbb{R}$.