Hello, Malkalypse!

If I interpret the problem correctly, angle z = angle CBA.

I assume that the $\displaystyle 14.776,\:14.967,\;x = 4.848$ are fixed

. . and that $\displaystyle y$ may vary. Code:

B
*
* *|
* z * |
14.776 * * |
* * |
* * |
* * |
C * * |y
* * |
* *_____ |
* *√x²+y² |
14.967 * * |
* * |
* * |
*- - - - - - - *
A x O

In right triangle $\displaystyle AOB$, we have: .$\displaystyle x = AO,\:y = BO$

. . Hence: .$\displaystyle AB = \sqrt{x^2+y^2}$

From the Law of Cosines: .$\displaystyle \cos z \;=\;\frac{CB^2 + AB^2 - CA^2}{2\cdot CB\cdot AB}$

. . so we have: .$\displaystyle \cos z \;=\;\frac{14.776^2 + (\sqrt{x^2+y^2})^2 - 14.967^2}{2(14.776)(\sqrt{x^2+y^2})} $

Since $\displaystyle x = 4.848$

. . we have: .$\displaystyle \cos z \;=\;\frac{y^2 - 17.822191}{29.552\sqrt{y^2 + 23.503104}} $