Obtuse angle is more than right

We are essentially looking for $P(x^2\ge 1-y^2 \mid x+y \ge 1)$ ; the conditioning holds since we already know that $x, y$ and $1$ form a triangle (triangle inequality).

Hence our answer is twice the area of the 2-d region bounded by $x+y=1$ and $x^2+y^2=1$ in the positive quadrant, which is,

$2\int_0^1 \int_{1-y}^{\sqrt{1-y^2}}\mathrm{d}x\mathrm{d}y=\pi/2-1$