Let $L_1$ and $L_2$ be two intersecting line segments. Let $\theta$ be the smaller angle of intersection between the two. $L_1$ is divided into segments of length $a$ and $b$ while $L_2$ is divided into segments of length $c$ and $d$ by the point of their intersection (call it $O$). A point $X$ is taken on $L_1$ with uniform probability across its length and another one, $Y$ on $L_2$ with the same probability distribution. Let $Z$ be the distance between these two points. How do we calculate the probability density function for $Z$. From what I can figure out, if $X$ and $Y$ are such that $\angle XOY$ is acute, then $Z^2 = OX^2 + OY^2 - 2*OX*OY*cos\theta$ and otherwise, $Z^2 = OX^2 + OY^2 + 2*OX*OY*cos\theta$. So perhaps the problem can be broken down into two parts.
$P(Z^2 > z) = P(Z^2 > z | \angle XOY acute) * P(\angle XOY acute) + P(Z^2 > z | \angle XOY obtuse) * P(~ \angle XOY obtuse)$.
I can't really understand how do I proceed further or if there is a simpler way from scratch.
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