Podem determinar que:
$$\begin{gathered} x_{1}=l_{1} \cos (\theta_1) ; \quad y_{1}=l_{1} \sin (\theta_1) ; \\ x_{2}=l_{2} \cos (\theta_1+\theta_2) ; \quad y_{2}=l_{2} \sin (\theta_1+\theta_2) \\ x=x_{1}+x_{2}=l_{1} \cos (\theta_1)+l_{2} \cos (\theta_1+\theta_2) \\ y=y_{1}+y_{2}=l_{1} \sin (\theta_1)+l_{2} \sin (\theta_1+\theta_2) \end{gathered}$$
Aquesta part és més difícil, perquè necessitem la cinemàtica directa per convertir-la de nou.
$$\begin{gathered} x^{2}+y^{2}=l_{1}^{2}+l_{2}^{2}+2 l_{1} l_{2} \cos (\theta_2) \\ \theta_{2}=\arccos \left(\frac{x^{2}+y^{2}-l_{1}^{2}-l_{2}^{2}}{2 l_{1} l_{2}}\right) \\ x=l_{1} \cos (\theta_1)+l_{2} \cos (\theta_1) \cos (\theta_2)-l_{2} \sin (\theta_1) \sin (\theta_2) \\ x=\cos (\theta_1)(l_{1}+l_{2} \cos (\theta_2))-l_{2} \sin (\theta_1) \sin (\theta_2) \\ y=l_{1} \sin (\theta_1)+l_{2} \sin (\theta_1) \cos (\theta_2)+l_{2} \cos (\theta_1) \sin (\theta_2) \\ y=\sin (\theta_1)(l_{1}+l_{2} \cos (\theta_2))+l_{2} \cos (\theta_1) \sin (\theta_2) \\ x+y=(l_{1}+l_{2} \cos (\theta_2))(\cos (\theta_1)+\sin (\theta_1))+l_{2} \sin (\theta_2)(\cos (\theta_1)-\sin (\theta_1)) \\ x-y=(l_{1}+l_{2} \cos (\theta_2))(\cos (\theta_1)-\sin (\theta_1))+l_{2} \sin (\theta_2)(\cos (\theta_1)+\sin (\theta_1)) \\ x+y+x-y=2x=2(l_{1}+l_{2} \cos (\theta_2)) \cos (\theta_1)+2 l_{2} \sin (\theta_2) \cos (\theta_1) \\ 2x=\cos (\theta_1)(2(l_{1}+l_{2} \cos (\theta_2))+2 l_{2} \sin (\theta_2)) \end{gathered}$$
$$\theta_{1}=\arccos \left(\frac{2x}{2(l_{1}+l_{2} \cos (\theta_2))+2 l_{2} \sin (\theta_2)}\right)$$