The angle in an isosceles triangle is a continuous uniform distribution in $(0,\pi)$ and between the two sides with the same length. Find the CDF of the perimeter of such a triangle.
I'm new to probability theory so bear with me.
First I need to express the perimeter with angle $\lambda$ and non-base-side $a$.
Using law of cosine I get $P=2a+c=2a+\sqrt{2a^2-2a^2\cos(\lambda)}$. Since base $c\in (0, 2a)$, the perimeter is in $(2a,4a)$.
Therefore: $F_U(z)=\mathbb{P}(U\leq z)=\mathbb{P}(\lambda\leq \cos^{-1}(\frac{-2a^2+4az-z^2}{2a^2})/\pi)$
The $\pi$ in the denominator must be there because it's a uniform distribution.
All in all, I get... $F_U=\begin{cases} 0 \,& z \leq 2a\\ \cos^{-1}(\frac{-2a^2+4az-z^2}{2a^2})/\pi \,& 2a< z< 4a \\ 1 \, & z\geq4a\end{cases}$
Am I correct? What could I do better (especially in notation)?