I understand how to find the surface area of a regular rhombicosidodecahedron. The formula given on Wikipedia is simple to use. Surface area is simply the sum of the areas of each face, which I can easily derive from plane trigonometry: $$ \begin{multline} A = 20\left(\frac12\cdot3\cdot\frac12 \cdot \tan{\left(\frac{\pi}{2}-\frac{\pi}{3}\right)}\right)a^2 \\ + 30\left(\frac12\cdot4\cdot\frac12 \cdot \tan{\left(\frac{\pi}{2}-\frac{\pi}{4}\right)}\right)a^2 \\ + 12\left(\frac12\cdot5\cdot\frac12 \cdot \tan{\left(\frac{\pi}{2}-\frac{\pi}{5}\right)}\right)a^2 \end{multline} $$
which simplifies to $$ A = \left(30+5\sqrt3+3\sqrt{25+10\sqrt5}\right)a^2 $$
But how would I go about deriving the formula for the volume of a regular rhombicosidodecahedron? On Wikipedia it is given as the following: $$ V = \frac{60+29\sqrt5}{3}a^3 $$
My thought process was, find the coordinates of each vertex in 3D space given an edge length $a$, assuming it is centered on the origin, then calculate the volume of each pyramid formed with the center by each face. But then that means I would have to know how to find the coordinates of each vertex, which I do not know how to do.
How then would I find those coordinates? Or, if that's a needlessly complex or even incorrect way to go about it, how do I derive this formula then?