Let's try to derive something. Assume the x-axis points to the right and the y-axis points up. I'm gonna use the subscripts b and s for bottom and side. The incoming velocity vector is decomposed in two components along the axes, so $v_s=v\sin\theta$.
Writing the total drag force as a vector:
$\vec D=\begin{bmatrix}-D_s\\D_b\end{bmatrix}=\begin{bmatrix}-C_{D_s}\frac{1}{2}\rho S v_s^2\\C_{D_b}\frac{1}{2}\rho S v_b^2\end{bmatrix}$
Taking the magnitude of the vector:
$D=C_D\frac{1}{2}\rho S v^2 = \sqrt{{C_{D_s}}^2\left(\frac{1}{2}\rho S\right)^2 v_s^4+ {C_{D_b}}^2\left(\frac{1}{2}\rho S\right)^2 v_b^4}$
This can be further simplified to:
$C_D = \sqrt{{C_{D_s}}^2\sin^4\theta+{C_{D_b}}^2\cos^4\theta}$
I don't think this can be simplified further.
Edit: S in this case is of course just the area of the rectangle, I got confused before