Great inverted disnub dodecahedron

The great inverted disnub dodecahedron, gidasid, or compound of twelve pentagrammic retroprisms is a uniform polyhedron compound. It consists of 120 triangles and 24 pentagrams (whcih fall in pairs in the same plane and combine into 12 stellated decagrams), with one pentagram and three triangles joining at a vertex.

This compound has rotational freedom, represented by an angle θ. we start at θ = 0° with all the pentagrammic retroprisms inscribed in a great icosahedron, and rotate pairs of retroprisms in opposite directions. At θ = 36° the retroprisms coincide by pairs, resulting in a double cover of the great inverted snub dodecahedron.

Vertex coordinates
The vertices of a great inverted disnub dodecahedron of edge length 1 and rotation angle θ are given by all even permutations of:
 * $$\left(±\frac{\sqrt5+(5-\sqrt5)\cos(\theta)}{10},\,±\sqrt{\frac{5-\sqrt5}{10}}\sin(\theta),\,±\frac{5-\sqrt5-4\sqrt5\cos(\theta)}{20}\right),$$
 * $$\left(±\frac{2\sqrt5+2\sqrt5\cos(\theta)+(1-\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{-10(1-\sqrt5)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta))}{20},\,±\frac{5-\sqrt5-(5+\sqrt5)\cos(\theta)-2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),$$
 * $$\left(±\frac{2\sqrt5-(5-3\sqrt5)\cos(\theta)+2\sqrt{\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{-10\cos(\theta)+(1-\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{5-\sqrt5+(5-\sqrt5)\cos(\theta)+(1+\sqrtt4)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),$$
 * $$\left(±\frac{2\sqrt5-(5-3\sqrt5)\cos(\theta)-2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{10\cos(\theta)+(\sqrt5-1)\sqrt{5\frac{5-\sqrt5}}{2}\sin(\theta)}{20},\,±\frac{5-\sqrt5+(5-\sqrt5)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right),$$
 * $$\left(±\frac{2\sqrt5+2\sqrt5\cos(\theta)-(\sqrt5-1)\sqrt{5\frac{5-\sqrt5}}{2}\sin(\theta)}{20},\,±\frac{10(\sqrt5-1)\cos(\theta)+(1+\sqrt5)\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20},\,±\frac{5-\sqrt5-(5+\sqrt5)\cos(\theta)+2\sqrt{5\frac{5-\sqrt5}{2}}\sin(\theta)}{20}\right).$$