How to find angle at centroid of triangle by its edge lengths?

Question

I need to write a program that takes length of a triangle's edges and calculates the angle $\angle APB$ ($P$ is the centroid of the triangle). Thanks for any help or clue.

Answer 1

$$ PA^2=\frac {2b^2+2c^2-a^2}9,\quad PB^2=\frac {2a^2+2c^2-b^2}9\\ \implies \cos(\angle APB)=\frac {PA^2+PB^2-c^2}{2\cdot PA \cdot PB} =\frac {a^2+b^2-5c^2}{2\sqrt{(2b^2+2c^2-a^2)(2a^2+2c^2-b^2)}}. $$