Obtuse the right way

The uniform sample space involved here is the unit square with lower left corner $(0,0)$ and upper right corner $(1,1).$

Now by the triangle inequality we know that in order for a triangle to be formed at all we require that $(x + y) \ge 1.$

Next, for a triangle to be obtuse we will require that $(x^{2} + y^{2}) \le 1,$ (as $x,y \le 1$ ).

The desired probability is then the area of the region of overlap of these two inequalities, which will be that of a quarter-circle of radius $1$ minus that of a right, isosceles triangle of side length $1,$ i.e., $\dfrac{\pi}{4} - \dfrac{1}{2} = \boxed{0.285}$ to 3 decimal places.

Let's take a more rigorous approach here! By the Law of Cosines we desire $1 = x^2 + y^2 - 2xy \cdot cos(\theta)$ , and solving for $\theta$ gives:

$cos(\theta) = \frac{x^2 + y^2 - 1}{2xy}$ (i)

and if we require an obtuse triangle, then:

$-1 < \frac{x^2 + y^2 - 1}{2xy} < 0$

which results in: $x^2 + y^2 < 1 \Rightarrow \boxed{y < \sqrt{1-x^2}}$ AND $-2xy < x^2 + y^2 - 1 \Rightarrow 1 < (x+y)^2 \Rightarrow \boxed{1 < x+y}$

of which the latter satisfies the Triangle Inequality. If $x,y$ are each independent & uniformly distributed over $[0,1]$ , then their pdf's both equal $f_{X}(x) = f_{Y}(y) = \frac{1}{1-0} = 1$ . Finally, the required probability we seek:

$P = \int_{0}^{1} \int_{1-x}^{\sqrt{1-x^2}} f_{X}(x) \cdot f_{Y}(y) dy dx = \int_{0}^{1} \int_{1-x}^{\sqrt{1-x^2}} dy dx = \int_{0}^{1} \sqrt{1-x^2} -1 + x dx$ ;

or $\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}arcsin(x) - x + \frac{x^2}{2}|_{0}^{1}$ ;

or $\boxed{\frac{\pi}{4} - \frac{1}{2}}.$