It looks like you have computed the "tetrahedra shapes" for the three ideal hyperbolic tetrahedra making up the knot complement.  Each of these gives a (euclidian!) shape to the four "cusp triangles" that "cut off" the four ideal vertices of the tetrahedron. To compute the shape of the cusp, you need to understand how it is tiled by the (in total 12) cusp triangles.  To do this by hand takes some work!  To get started, draw one of the tetrahedra $t$ in the upper-half-space model (with a vertex at infinity), and consider how the other three tetrahedra, that also meet infinity, and that also meet $t$, lie in the model.  Each of these has just one cusp triangle that separates the body of the tetrahedron from infinity. You want to use the shapes $x_i$, $y_i$, and $z_i$ to tile these outward... (say on the horoplane at height one).