What kind of triangle is formed by three unequal masses in a circular restricted three body orbit?
These questions are answered by the same references listed in my previous answer.

In the Newtonian limit, an equilateral 3-body solution exists for any combination of masses. (However, it is stable only if the inequality in the previous answer is satisfied.). This equilateral configuration orbits the center of mass, which is generally not in the center of the equilateral triangle. Each of the bodies follows a circular orbit with radius (see e.g. 1212.0754

$$ r_1 = a\sqrt{\nu_2^2+\nu_2\nu_3+\nu_3^2}$$
$$ r_2 = a\sqrt{\nu_1^2+\nu_1\nu_3+\nu_3^2}$$
$$ r_3 = a\sqrt{\nu_1^2+\nu_1\nu_2+\nu_2^2}$$

and frequency

$$\omega = \sqrt{M/a^3}, $$

where $a$ is the length of the sides of the equilateral triangle, $M$ the total mass, and $\nu_i = m_i/M$.

The situation changes when one accounts for relativisitic effects. When one accounts for the leading (post-Newtonian) corrections, then  a circular restricted three body solution still exists for general masses (with a smaller region of stability than in the Newtonian case.). However, the triangular configuration is no-longer circular (unless all three mass are equal or two masses are 0). Keeping the distances to the center of mass $r_i$ as in the Newtonian case, the sides of the triangle are now given by (again see  1212.0754)

$$ r_{ij} = a(1+\frac{M}{a}\epsilon_{ij}+ O(\tfrac{M^2}{a^2}) )$$

with

$$\epsilon_{12} = \frac{1}{24}[(\nu_1-\nu_3)(5-3\nu_2)+(\nu_2-\nu_3)(5-3\nu_1)]$$
$$\epsilon_{23} = \frac{1}{24}[(\nu_2-\nu_1)(5-3\nu_3)+(\nu_3-\nu_1)(5-3\nu_2)]$$
$$\epsilon_{31} = \frac{1}{24}[(\nu_3-\nu_2)(5-3\nu_1)+(\nu_1-\nu_2)(5-3\nu_3)].$$

The angles of the triangle can be calculated from these lengths if one is so inclined.