I have a practical problem (let's call it problem $P_1$) like the following:
there are three vectors $(x_1, x_2) \in \mathbb{R}^2$, $(x_1,x_2) = (r_x\cos{\theta_x}, r_x\sin{\theta_x})$, $(y_1, y_2) \in \mathbb{R}^2$, $(y_1,y_2) = (r_y\cos{\theta_y}, r_y\sin{\theta_y})$ and $(z_1, 0) \in \mathbb{R}^2$ such that:
$$(y_1, y_2) = (x_1,x_2) + (z_1,0)$$
$r_x$ and $r_y$ are known. I know also $\alpha_x = \theta_x + \beta$ and $\alpha_y = \theta_y + \beta$. Values of $\beta$ and $z_1$ are unknown and must be determined.
I have a solution for this problem:
$$z_1 = \sqrt{r_x^2+r_y^2-2r_xr_y\cos{(\alpha_x-\alpha_y)}}$$
$$\beta = \alpha_x – atan2\left(\frac{\sin(\alpha_x-\alpha_y)}{\cos(\alpha_x-\alpha_y)-\frac{r_x}{r_y}}\right)$$
Now suppose that I have a similar problem $P_2$:
there are five vectors $(x_1, x_2) \in \mathbb{R}^2$, $(x_1,x_2) = (r_x\cos{\theta_x}, r_x\sin{\theta_x})$, $(y_1, y_2) \in \mathbb{R}^2$, $(y_1,y_2) = (r_y\cos{\theta_y}, r_y\sin{\theta_y})$, $(z_1, z_2) \in \mathbb{R}^2$, $(z_1,z_2) = (r_z\cos{\theta_z}, r_z\sin{\theta_z})$, $(t_1, t_2) \in \mathbb{R}^2$, $(t_1,t_2) = (r_t\cos{\theta_t}, r_t\sin{\theta_t})$, $(u_1, u_2) \in \mathbb{R}^2$, $(u_1,u_2) = (r_u\cos{\theta_u}, r_u\sin{\theta_u})$ such that:
$$(y_1, y_2) = (x_1,x_2) + (z_1,z_2)$$
$$(t_1, t_2) = (x_1,x_2) + (u_1,u_2)$$
$r_y$, and $r_t$ are known. I know also $\alpha_y = \theta_y + \beta$, $\alpha_t = \theta_t + \beta$, $\theta_z$, $\theta_u$, ${r_z}/{r_u}$. Values of $\beta$, $r_x$, $r_z$, $r_u$ and $\theta_x$ are unknown and must be determined. Let's say that I can be satisfied by determining only $\beta$, $\theta_x$ and ${r_x}/{r_z}$ (let's call this relaxed problem $P_3$).
Is there a solution to $P_2$ or $P_3$?
I believe there is no solution, because of the difficulty of finding $\beta$ without any "absolute reference" vector like $(z_1, 0)$ in the first $P_1$ problem (where $r_x$ and $\alpha_x$ are known, like if we had $(u_1, u_2) = (0,0)$ in $P_2$), but I could be wrong.
Also, if the answer is negative, would it be possible to determine all unknowns adding another equation $(v_1,v_2)=(x_1,x_2)+(w_1,w_2)$ where the magnitude of $(v_1,v_2)$ is known and its phase also but up to an offset $\beta$, all alike the first two equations?
Best Answer
Let $O$ be the origin, and let $\vec{OX}=(x_1,x_2),\vec{OY}=(y_1,y_2)$ and $\vec{OT}=(t_1,t_2)$.
Applying the law of cosines to $\triangle{OYT}$, one gets $$YT=\sqrt{r_y^2+r_t^2-2r_yr_t\cos(\alpha_y-\alpha_t)}$$
Applying the law of cosines to $\triangle{XYT}$, one gets $$YT^2=r_z^2+r_u^2-2r_zr_u\cos(\theta_z-\theta_u)$$ which can be written as $$YT^2=\bigg(\bigg(\frac{r_z}{r_u}\bigg)^2+1-2\bigg(\frac{r_z}{r_u}\bigg)\cos(\theta_z-\theta_u)\bigg)r_u^2$$ so one gets $$r_u=\frac{\sqrt{r_y^2+r_t^2-2r_yr_t\cos(\alpha_y-\alpha_t)}}{\sqrt{\bigg(\dfrac{r_z}{r_u}\bigg)^2+1-2\bigg(\dfrac{r_z}{r_u}\bigg)\cos(\theta_z-\theta_u)}}$$ and $$r_z=\dfrac{r_z}{r_u}\times r_u=\frac{r_z}{r_u}\cdot\frac{\sqrt{r_y^2+r_t^2-2r_yr_t\cos(\alpha_y-\alpha_t)}}{\sqrt{\bigg(\dfrac{r_z}{r_u}\bigg)^2+1-2\bigg(\dfrac{r_z}{r_u}\bigg)\cos(\theta_z-\theta_u)}}$$
One has $$r_y\cos\theta_y=r_x\cos\theta_x+r_z\cos\theta_z\tag1$$ $$r_y\sin\theta_y=r_x\sin\theta_x+r_z\sin\theta_z\tag2$$ $$r_t\cos\theta_t=r_x\cos\theta_x+r_u\cos\theta_u\tag3$$ $$r_t\sin\theta_t=r_x\sin\theta_x+r_u\sin\theta_u\tag4$$ It follows from $(1)-(3)$ and $(2)-(4)$ that $$\cos\theta_y=\frac{K+r_t\cos\theta_t}{r_y},\qquad \sin\theta_y=\frac{L+r_t\sin\theta_t}{r_y}$$ where $$K=r_z\cos\theta_z-r_u\cos\theta_u,\qquad L=r_z\sin\theta_z-r_u\sin\theta_u$$ Hence, one gets $$1=\cos^2\theta_y+\sin^2\theta_y=\bigg(\frac{K+r_t\cos\theta_t}{r_y}\bigg)^2+\bigg(\frac{L+r_t\sin\theta_t}{r_y}\bigg)^2$$ which can be written as
$$(K+r_t\cos\theta_t)^2+(L+r_t\sin\theta_t)^2=r_y^2$$ $$K^2+r_t^2+2Kr_t\cos\theta_t+L^2+2Lr_t\sin\theta_t=r_y^2$$ $$2Kr_t\cos\theta_t+2Lr_t\sin\theta_t=r_y^2-K^2-L^2-r_t^2$$ $$\sqrt{(2Kr_t)^2+(2Lr_t)^2}\cos(\theta_t-p)=r_y^2-K^2-L^2-r_t^2$$ where $$\cos p=\frac{2Kr_t}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}},\sin p=\frac{2Lr_t}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}}$$ So, $$\theta_t=p+\arccos\bigg(\frac{r_y^2-K^2-L^2-r_t^2}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}}\bigg)$$ $$\beta=\alpha_t-\theta_t=\alpha_t-p-\arccos\bigg(\frac{r_y^2-K^2-L^2-r_t^2}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}}\bigg)$$ $$\theta_y=\alpha_y-\beta=\alpha_y-\alpha_t+p+\arccos\bigg(\frac{r_y^2-K^2-L^2-r_t^2}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}}\bigg)$$
It follows from $(1)(2)$ that $$r_x=\sqrt{(r_x\cos\theta_x)^2+(r_x\sin\theta_x)^2}$$ $$=\sqrt{(r_y\cos\theta_y-r_z\cos\theta_z)^2+(r_y\sin\theta_y-r_z\sin\theta_z)^2}\tag4$$
Finally, $\theta_x$ can be determined by $(1)(2)(4)$.
Added :
We have already got $r_u$ and $r_z$.
Also, we have already seen that $$(1)(2)(3)(4)\implies \cos(\theta_t-p)=M\quad \text{where}\quad M=\frac{r_y^2-K^2-L^2-r_t^2}{\sqrt{(2Kr_t)^2+(2Lr_t)^2}}$$ from which one gets $$\theta_t=p\color{red}{\pm}\arccos(M)$$ where $\arccos (M)$ is such that $$\cos(\arccos(M))=M\qquad\text{and}\qquad \color{red}{0\leqslant\arccos(M)\leqslant \pi}$$
So, one gets $$\beta=\alpha_t-\theta_t=\alpha_t-p\mp\arccos(M)$$ and $$\theta_y=\alpha_y-\beta=\alpha_y-\alpha_t+p\pm\arccos(M)$$
It follows from these that $$(1)(2)(3)(4)$$ $$\implies (\theta_t,\beta,\theta_y)=(p+\arccos(M),\alpha_t-p-\arccos(M),\alpha_y-\alpha_t+p+\arccos(M)),(p-\arccos(M),\alpha_t-p+\arccos(M),\alpha_y-\alpha_t+p-\arccos(M))$$
For sufficiency, one finally has to check if each of these satisfies $(1)(2)(3)(4)$.