An acute triangle from a broken stick

Idea. The triangle fails to be acute when one piece is too long: $c^2\ge a^2+b^2$ for one of the pieces. At most one piece can be that long, and by symmetry each piece is the long one with the same probability. So compute one probability and subtract three times it from 1.

Step 1: the pieces. The three lengths $(a,b,c)$ are uniform on the set $a,b,c>0$, $a+b+c=1$; equivalently $(a,b)$ is uniform on the triangle $a,b>0$, $a+b<1$, with density 2, and $c=1-a-b$.

Step 2: one bad event. $c^2>a^2+b^2$ with $c=1-a-b$ becomes $$1-2a-2b+2ab>0⟺b<\frac{1-2a}{2(1-a)},$$ which requires $a<\frac{1}{2}$. Its probability is $$\begin{array}{rl}2{\int }_0^{1/2}\frac{1-2a}{2(1-a)}da& ={\int }_{1/2}^1\frac{2u-1}{u}du\\ & =1-\ln 2,\end{array}$$ with $u=1-a$.

Step 3: three bad events. Two of them cannot hold together (if $c^2\ge a^2+b^2$ then $c$ is the longest piece). So $$\mathbb{P}(\text{acute})=1-3(1-\ln 2)=3\ln 2-2,$$ about 0.079.

Compare. The pieces form a triangle with probability $\frac{1}{4}$. So only about a third of those triangles are acute; an obtuse triangle comes out with probability $\frac{9}{4}-3\ln 2\approx 0.171$.