I have a chain of line segments (vectors also work) such that every line segment begins at the end of another, save for the first which begins at a given point $P_0$, such as in the picture (1), and are of a known length $L_n$, where n corresponds to the number of the line segment and L is the length of said line segment or magnitude of the vector.
Such a system follows the rule $$P = \left(\sum_{n=1}^{N}L_n\cos\left(\theta_n\right),\sum_{n=1}^{N}L_n\sin\left(\theta_n\right)\right)$$
where $N$ is the number of line segments and $P$ is the desired point.
I have already found that for $N = 2$ members the solution of the system
$$ L_1\cos(\theta_1) + L_2\cos(\theta_2) =P_x $$
$$ L_1\sin(\theta_1) + L_2\sin(\theta_2) =P_y $$
give $\theta_1$ and $\theta_2$ relatively easily as long as $L_1+L_2 > P_x + P_y $, in my experience.
Is there a way to solve for the angles (relative to the x-axis) when $N > 2$? My attempts have been fruitful so far, yet difficult. This was the last attempt I made for $N=3$, is the end of the line segment, like where an arrow would fall on a vector. (please excuse the lack of braces, they are not cooperating):
First equation is solved for $\theta_3$, $P_{2_{x}}$, and $P_{2_{y}}$:
$$ P_{2_{x}}+L_3\cos(\theta_3)=P_x $$
$$ P_{2_{y}}+L_3\sin(\theta_3)=P_y $$
$$\sqrt{({L_3\cos(\theta_3)})^2+({L_3\sin(\theta_3)})^2} = L3$$
the second system uses those to find the values of $\theta_2$, $P_{1_{x}}$, and $P_{1_{y}}$ $$ P_{1_{x}}+L_2\cos(\theta_2)=P_{2_{x}} $$
$$ P_{1_{y}}+L_2\sin(\theta_2)=P_{2_{y}} $$
$$\sqrt{({L_2\cos(\theta_2)})^2+({L_2\sin(\theta_2)})^2} = L3$$
then working through the third finishes the problem by finding the values needed from the starting point. $$ P_{0_{x}}+L_1\cos\left(\arctan\left(\frac{P_{1_{y}}}{P_{1_{x}}}\right)\right)=P_{1_{x}} $$
$$ P_{0_{y}}+L_1\cos\left(\arctan\left(\frac{P_{1_{y}}}{P_{1_{x}}}\right)\right)=P_{1_{y}} $$ The final one is specifications of the starting point (Origin is assumed).
$$P_{0_{x}}=0$$ $$P_{0_{y}}=0$$
These all come together to the system below, which corresponds to the summation above;
$$P_{0_x}+L_1\cos\left(\arctan\left(\frac{P_{1_{y}}}{P_{1_{x}}}\right)\right) + L_2\cos(\theta_2)+L_3\cos(\theta_3)=P_{x}$$
$$P_{0_y}+L_1\sin\left(\arctan\left(\frac{P_{1_{y}}}{P_{1_{x}}}\right)\right) + L_2\sin(\theta_2)+L_3\sin(\theta_3)=P_{y}$$
Is there a simpler way to express this idea or solve for all $\theta_n$ for $N$ number of segments of known length $L$? It is quite exhausting having to solve three systems in a row. I hope you have understood what I have written here. Thank you.