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I have reduced a problem I'm trying to solve, to the following $4$ equations:

$$\cos(\theta) = \frac{A}{b+c}$$ $$\sin(\theta) = \frac{D-e}{b-c}$$ $$\sin(\theta) = \frac{c}{e}$$ $$\cos(\theta)^2 + \sin(\theta)^2 = 1$$ where only $A$ and $D$ are known.

I having a hard time solving this analytically. Is there an analytic solution to these equations or can a solution only be found numerically? Is there a(n) (easy) way to tell whether it is one or the other?

Jens
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    You only have 2 equations... The last equation is always satisfied and $\theta$ can be eliminated. In the end you just have $\frac ce = \frac{D-e}{b-c}$ and $\frac{c^2}{e^2} + \frac{A^2}{(b+c)^2}=1$. – PierreCarre Feb 01 '19 at 23:11
  • Thanks @JohnOmielan, i'll correct the comment. – PierreCarre Feb 01 '19 at 23:16
  • @PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case? – Jens Feb 01 '19 at 23:26
  • The last equation is an universal condition. You may as well consider $0=0$ as your last equation. – PierreCarre Feb 01 '19 at 23:27
  • Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $\sin(\theta),,$ and $,c,$ the solution of a linear equation in $,b,\sin(\theta),$ and $,e,$ is linear in $,b,c,\sin(\theta).,$? – Somos Feb 02 '19 at 00:19
  • Do you mean this problem? It’s helpful to potential answers to provide a pointer to it for context. – amd Feb 02 '19 at 00:43

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