Here's the progress I (think I) have made.
EQUILATERAL TRIANGLES
For Equilateral triangles, the probability is zero for any angle.
ISOSCELES TRIANGLES
For Isosceles triangles, when defining one of the equal sides as the radii of a circle, any other radii could possibly be the other equal side of the triangle and the two radii will have an obtuse angle only for half the central circular angle.
And as far as the base angles are concerned, they cannot be obtuse as two of them would make an angle greater than 180, more than what our triangle can have.
Since the angle chosen is one of the three possible choices which are equally likely, the probability of an obtuse angle in a triangle is 1/6.
SCALENE TRIANGLES
For Scalene triangles, we can apply a similar technique where we would have two intersecting circles of equal radius with their center lying of the circumference of the other circle. In that case the radii joining the two center points could be defined as the longest side of the scalene triangle and the third point lying of the edge of the intersection of the circles or on the line joining the two vertices of the intersectional area of the circle would create either isosceles or equilateral triangles (equilateral if the third point is at the vertex, else isosceles).
And if we define a circle with its circumference passing through the two already established center points as diametrically opposite points, we will get a right angle when the third point is on the circumference of the smaller circle now formed (Thales Theorem, right angle in a semicircle). When lying inside this circle, the third point thus creates a scalene triangle given that it does not lie in the middle line joining the two vertices of the intersection.
And since the third point cannot create an obtuse angle using the base side (radii of the big equal circles) while lying in the intersection (maximum can be 60 degrees), the ratio of the area of the small circle to the area of the intersection gives us our required probability of obtuse angle in a scalene triangle to be $\frac{\pi}{8 \pi/3 - 4 \sqrt{3}}$.
Now I am stuck at two things:
- How to take into account all those points which would make up an equilateral or an isosceles triangle in case of scalene triangle (I assumed them to negligibly affect the probability as they are much less in number as compared to the area of the intersection).
- What is the probability of the triangle to be equilateral, isosceles and scalene.
From intuition and also comparison of the possible third points making up equilateral, isosceles and scalene triangles, we can say that the three types of triangles are not equally probable.
How should we continue?
As a side note I was wondering how the same thing reacts to curved spaces but am finding it difficult so I would appreciate insights on that as well.
As indicated in comments by both Dan Asimov and myself, the question is not sufficiently well defined to produce a unique answer. At issue is the process by which a "random triangle" is generated.
In this sense, the problem is similar to the question posed by the Bertrand paradox: If a random chord is chosen, what is the probability that it's longer than the side of an equilateral triangle inscribed in that circle? Depending on how the chord is chosen, the answer is either $1/2$, $1/3$, or $1/4$. (I suspect other answers could also be justified.)
Here are two plausible processes by which a random triangle could be chosen. First, we might say that the angles of the triangle are more important to this problem than the sides. Since we know that the angles must add up to $180$ degrees, we randomly select, uniformly, two real numbers $a$ and $b$ between $0$ and $180$. Without loss of generality, let $a$ < $b$. Then we'll say that our candidate triangle has angles $a$, $b-a$, and $180-b$. You'll see that this is always a triangle, and all possible triangles are chosen (up to scaling). With this selection process, we get an obtuse angle if $a$ and $b$ are both less than $90$, if they're both greater than $90$, or if they're separated by at least $90$. This happens with probability $3/4$.
On the other hand, we might say that the sides are more important than the angles. In that case, we'll construct our triangle by breaking a unit stick in two points by uniformly selecting two real numbers $a$ and $b$ between $0$ and $1$. Again, without loss of generality, we'll say that $a < b$. Then we'll say that our triangle has sides $a$, $b-a$, and $1-b$. In addition, we'll have to reject any cases where the three lengths don't form a triangle: if $b < 1/2$, if $a > 1/2$, or if $b-a > 1/2$. With this selection process, creating a triangle with sides $x$, $y$, and $z$, we get an obtuse angle if either $x^2+y^2 < z^2$, $y^2+z^2 < x^2$, or $z^2+x^2 < y^2$. This happens with probability $9-12\ln 2 \approx 0.68223$.
At one level, the difference in the two answers comes about because if you choose the side lengths uniformly, you get a non-uniform distribution for the angles; on the other hand, if you choose the angles uniformly, you get a non-uniform distribution for the sides. In no way can you get them both to come out uniform.
I suspect that, as in the Bertrand paradox, there are other processes that lead to other defensible answers. So to answer this more definitively, we'll need to nail down what the process is by which the triangle is produced.