Given the two distributions of two sides of a triangle (for example, Uniform and Rayleigh) and the distribution of an angle between them (Uniform[0,Pi]), find the length of the third side.
What i came up with is: Using the cosine law, I get the formula for the third side of a triangle: $$ a^2 = b^2 + c^2 - 2bc\cos\alpha\ $$ However, the third term is dependent (is it?) on the first two. Next i found in a book the method for multivariate PDF transformation using jacobian and inverse function (even though in my case it splits in two domains). But there I came to a stop, as in the end I have to integrate over the second and third parameter in order to get a one dimensional distribution, and there i still fail to get the right boundaries.
I'll really appreciate any help or tips on this problem.
Edit: Forgot to mention that the three mentioned random variables are independent.
Edit2: Sorry for the long delay. Here's what I came up with: In short, I have to sum the next value for all branch cuts: $$W_{\eta_1}\left(y_1\right) = \int\limits_{y_2, y_3} w_{\xi^3}\left (\phi \left (y^3 \right ),y_2,y_3 \right ) \cdot \left |\frac{\partial \phi(y^3)}{\partial y_1} \right | dy_2 \; dy_3$$ However, the resulting integral is (as it might be seen) not that simple.
Is it the only way solve this problem, or are there another ways&
To find distribution of third side, $Z$, given distributions of two other sides, $X$ and $Y$, and angle $A$ it boils down to finding distribution of $L=Z^{2}=X^{2}+Y^{2}-2XY\cos(A)$. Since distribution of $Z$ is given by CDF $$F_{Z}(z)=P(Z\leq t)=P(Z^{2}\leq t^{2})=P(L\leq t^{2})$$ So finding CDF of L will help us find CDF of $Z$. But this can be quite messy because you have two calculate distribution of $X^{2}$, $Y^{2}$, and $2XY\cos(A)$ then distribution of $X^{2}+Y^{2}-2XY\cos(A)$. I have feeling tis won't come to a closed form. But if you wanted to get maybe a guess what it could be you could always simulate $X$, $Y$, and $A$ and use those observations to find simulations of $Z^{2}$ which give you some idea of what it may look like