probability of three random points inside a circle forming a right angle triangle three points are randomly chosen on a circle.
what the probability that
1.triangle formed is right angled triangle.
2.triangle formed is acute angled triangle.
3.triangle formed is obtuse angled triangle.
 A: The question is about points on a circle, so that's the one I'm answering here. The triangle is right angled precisely if two of the three points lie on a diameter of the circle.  This is Thales' theorem.  As already noted, the probability that this occurs is $0$.  The triangle is obtuse precisely if all three points lie on the same side of some diagonal of the circle.  The probability for that is $\frac{3}{4}$ (assuming a uniform distribution on the circle).
One way of seeing this is to introduce for each vertex $v_k$ a random variable $X_k$ as follows:
$$
X_k = \begin{cases}
1 & \textrm{if the other vertices lie on the half circle starting at } v_k \textrm{ in clockwise direction}\\
0 & \textrm{otherwise}
\end{cases}
$$
Then at most one of $X_1, X_2, X_3$ can be equal to one.  Therefore
$$
\mathbb{P}(\textrm{triangle is obtuse}) = \mathbb{P}(X_1+X_2+X_3 = 1) = \mathbb{E}(X_1+X_2+X_3) = 3 \mathbb{E}(X_1).
$$
The last equality follows from the fact that $X_1, X_2, X_3$ all have the same probability distribution. Now $\mathbb{E}(X_1) = \frac{1}{4}$ since both $v_2$ and $v_3$ have a probability of $\frac{1}{2}$ to lie on the half circle starting at $v_1$.
