I suggest, we construct a kind of "triangles generator":
We take a circle of any diameter, draw a chord AC from the endpoints of the chord we draw two line segments, the ends of which intersect on the circle.
Symmetrically we draw two other segments on the other side of the chord and we got two triangles.
When the chord is coincides with diameter the triangles are identical.
The final touch – we connect the vertices of the triangles by line a-a.
"Triangles generator" is ready let's start generate.
For this purpose we are doing two things:
1) Moving the chord parallel to itself from the diameter stepwise with any resolution you choose.
2) At each step we move the line a-a together with the vertices of the triangles from point B to point A.
By action 1 we iterate through all possible angles at the vertices of the triangles. (Look at nice animation here)
By action 2 - iterate through all possible conjugate base angles.
Eventually we generate all possible variants of triangles.
At the resolution tending to zero we get close to infinite number of triangles, but an interesting observation for us, that in any case each specific obtuse triangle always has one corresponding acute-angled triangle.
In other words:
THE PROBABILITY OBTAIN OBTUSE TRIANGLE AT RANDOM CHOICE IS 50%.
One must be careful when defining random triangles.
There are many ways, yielding various distributions.
The OP requires a particular method: selecting three points at random. = which means choose random triangle =
This is easily done (for the unit circle) by selecting x and y values in the closed
interval [-1, 1] and discarding any points for which x2+ y2>1.
Inspecting a million or so triangles constructed from triplets of such
points gives a clear answer to the probability in a finite region.,
Please pay attention - I prove clearly that any obtuse triangle corresponds to one of the acute-angled, which shows clearly that we have 50% chance to choose one or the type no matter in which way you do it.