This is just a system of linear equations, writing it as a matrix, \[ \begin{bmatrix} 0 &c &b\\ c &0 &a\\ b &a &0 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} a\\ b\\ c \end{bmatrix} \]By finding the inverse of the matrix, we get \begin{align*} \begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix} 0 &c &b\\ c &0 &a\\ b &a &0 \end{bmatrix} ^{-1} \begin{bmatrix} a\\ b\\ c \end{bmatrix}\\ &= \begin{bmatrix} -\frac{a^2}{2bc} &\frac{b}{2c} &\frac{c}{2b}\\ \frac{a}{2c} &-\frac{b^2}{2ca} &\frac{c}{2a}\\ \frac{a}{2b} &\frac{b}{2a} &-\frac{c^2}{2ab} \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix}\\ &= \begin{bmatrix} \frac{b^2+c^2-a^2}{2bc}\\ \frac{c^2+a^2-b^2}{2bc}\\ \frac{a^2+b^2-c^2}{2ab} \end{bmatrix} \end{align*}Which surprisingly resembles the cosine rule! Unfortunately nothing guarantees that $a$, $b$, $c$ are the lengths of the sides of a triangle. We can easily check that \[\frac{1-x^2}{a^2} = \frac{2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4+b^4+c^4)}{4a^2b^2c^2}\]and since this expression is symmetric, we have $\frac{1 - x^2}{a^2} = \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2}$.