I am trying to analytically solve these equations for the three variables of $\theta$, $L_p$, and $R_c$. Matlab can not solve them! I am wondering if there is any solution for this at all? And how I can find it?
$$ \left\{\begin{array}{rcl} R_c-R_c \cos \theta +L_p \sin \theta &=& \sqrt{X^2+Y^2} \\ R_c \sin \theta +L_p \cos \theta &=& Z \\ R_c &=& \frac{L}{\theta} \end{array} \right. $$
As Travis commented, there is no hope to obtain an analytical solution for this system of equations since $\theta$ appears alone and also involved in trigonometric functions (remember that we do not know the solution of the equation $x=\cos(x)$).
However, we can reduce the system to a single equation in $\theta$. Eliminating $R_c$ and $L_p$ from the first and second equations leads to $$R_c=\frac{\cos (\theta ) \sqrt{X^2+Y^2}-Z \sin (\theta )}{\cos (\theta )-1}$$ $$L_p=\cot \left(\frac{\theta }{2}\right) \sqrt{X^2+Y^2}-Z$$ Using the third equation then leads to $$f(\theta)=\frac{\cos (\theta ) \sqrt{X^2+Y^2}-Z \sin (\theta )}{\cos (\theta )-1}-\frac{L}{\theta}=0$$ in which all terms are known. Solve this last equation for $\theta$ and get $R_c$ and $L_p$ from the solutions.
In any manner, numerical methods, such as Newton, will be required but all of that is doable provided a good look at the graph of the function for getting reasonable estimates of the solution.
If the solutions do not correspond to $\cos (\theta )=1$ or $\theta=0$, it is probable that function $$g(\theta)=\cos (\theta ) \left(\theta \sqrt{X^2+Y^2}-L\right)+L-\theta Z \sin (\theta )$$ could be more pleasant.
If by chance, $\theta$ is small, a Taylor expansion of $g(\theta)$ built at $\theta=0$ gives an approximation $$\theta \approx \frac{2 \sqrt{X^2+Y^2}}{2 Z-L}$$