Three Points

Let $A$ , $B$ , and $C$ be the points on the circle and let $O$ be the center of the circle. Then without loss of generality, the points can be rotated so that $A$ is on the positive $x$ -axis. Then let $x$ be the central angle between $A$ and $B$ and $y$ be the central angle between $A$ and $C$ . Then both $x$ and $y$ are such that $0 \leq x < 360°$ and $0 \leq y < 360°$ .

$\triangle AOB$ , $\triangle BOC$ , and $\triangle AOC$ are isosceles triangles because each have two sides that are radii of the circle.

If $y > x$ , the base angles of these triangles are $90° - \frac{x}{2}$ , $90° - \frac{y}{2} + \frac{x}{2}$ , and $90° - \frac{y}{2}$ , which makes $\angle BAC = \frac{y - x}{2}$ , $\angle ABC = 180° - \frac{y}{2}$ , and $\angle ACB = \frac{x}{2}$ . For any one of these angles to be greater or equal to an obtuse angle $\alpha$ , either $\frac{y - x}{2} \geq \alpha$ or $y \geq x + 2\alpha$ , $180° - \frac{y}{2} \geq \alpha$ or $y \leq 360° - 2\alpha$ , or $\frac{x}{2} \geq \alpha$ or $x \geq 2\alpha$ , which is shaded in green in the graph below:

By symmetry, if $x > y$ , for any one of these angles to be greater or equal to an obtuse angle $\alpha$ , either $y \leq x - 2\alpha$ , $x \leq 360° - 2\alpha$ , or $y \geq 2\alpha$ , which is shaded in green in the graph below:

Combining these graphs gives a visual representation of all the possible triangles that can be made by three points on the circumference of a circle, with the green regions showing the triangles that have at least one angle that is greater or equal to an obtuse angle $\alpha$ :

The green regions have a combined area $A_G$ equivalent to $3$ squares with sides of $360° - 2\alpha$ , so $A_G = 3(360° - 2\alpha)^2$ , and the total area $A_T$ is a square with sides of $360°$ , so $A_T = 360°^2$ . The probability $P$ of the triangle having an obtuse angle that is greater or equal to $\alpha$ is therefore $P = \frac{A_G}{A_T} = \frac{3(360° - 2\alpha)^2}{360°^2}$ or $P = \frac{3(180° - \alpha)^2}{180°^2}$ , and when $P = \frac{1}{12}$ , this solves to $\alpha = \boxed{150°}$ .