The average angle formed by right triangles with a variable length adjacent side?

Question

Suppose we have a right triangle with an opposite edge of constant length $y$ and an adjacent edge whose length varies continuously between $x_1$ and $x_2$. Is there a way to find the average angle this triangle forms?

Answer 1

You have to specify the probability distribution of the adjacent edge, but then, yes there is. If $x$ is the length of the adjacent edge, the angle is $\arctan \left(\frac yx \right)$. The average angle is then $\int_{x_1}^{x_2} \arctan \left(\frac yx \right)p(x) dx$ where $p(x)$ is the probability distribution function of $x$ when constrained between $x_1$ and $x_2$. If the distribution is uniform, $p(x)=\frac 1{x_2-x_1}$. Alpha did not find a symbolic integral, so you are probably going to have to integrate it numerically.

Answer 2

Let $\triangle ABC$ be right angled at $B$ and let $BC=\ell$ (fixed length) and $AB=x$ (variable), where $a \leq x \leq b$ and let $\angle BCA=\theta$. Then $$\theta=\arctan \frac{\ell}{x}.$$ So the average value of $\theta$, when $x \in [a,b]$ will be given by $$\frac{1}{b-a}\int_a^b \theta \, dx==\frac{1}{b-a}\int_a^b \arctan \frac{\ell}{x} \, dx$$