As far as I understand your sketch ...
1. The vertex of the pyramid (Obviously you labeled it V) is perpendicularly above the intersection M of the two diagonals of the base area.
The diagonal has a length of 25.
2. The height h, a half diagonal and an edge form a right triangle:
$\displaystyle 24^2 = 12.5^2+h^2$
Solve for h.
3. The angle $\displaystyle \angle(MVC)$ is half as large as the angle $\displaystyle \angle(AVC)$. Use the right traingle MCV to determine the angle $\displaystyle \angle(MVC)$:
$\displaystyle \sin(\angle(MVC)) = \dfrac {12.5}{24}~\implies~\angle(MVC)\approx 31.388^\circ$
1. You should have seen that $\displaystyle |\angle(CAV)| = |\angle(VCM)|$.
2. Let d denote the length of one diagonal of the base area and e the length of one edge, then you see that
$\displaystyle \cos(\angle(VCM)) = \dfrac{\frac12 d}{e}$.
3. Calculate the value of $\displaystyle \angle(VCM)$ and subsequently the value of $\displaystyle \angle(CAV)$.
As I've suggested in the post #8 you'll get the Cosine of $\displaystyle \angle(VCM)$ by calculating
$\displaystyle \cos(\angle(VCM)) = \dfrac{\frac12 d}{e}$
$\displaystyle \cos(\angle(VCM)) = \dfrac{12.5}{24} = \dfrac{25}{48}$
And now use the inverse Cosine function of your calculator.
The final result should be $\displaystyle \angle(VCM) \approx 58.61^\circ$