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I have been mulling over a really interesting question in analytic geometry that is much harder than it first appears to be. Hope you can provide some insight into solving it.

Diagram of Problem

If you only know:

  1. The lengths of line segments A, B and C
  2. The coordinates of the end position of line segment C (X,Y)

Define the internal angles d, e and f in terms of A, B, C and (X,Y).

What I've done to approach the problem:

  1. Make a right angled triangle from origin, (X,Y) and (X,0) {last point is simply the where (X,Y) meets the x axis}
  2. Draw an imaginary line from where line segments A and B meet to (X,Y)
  3. Define angle between A and hypotenuse of the right triangle from (1) using the imaginary line in (2) through cosine rule
  4. Define angle f using the imaginary line in (2) through cosine rule
  5. Equate the two equations in (3) and (4) so that the two unknown angles are defined in terms of each other

I am stuck at this point. Am I following the correct approach, or does something else need to be done to define angles d, e and f using only information provided? Thanks!

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Are you allowed to define the angles in terms of any trig function of the line segments? – homegrown Mar 4 '14 at 20:22
Yes you are! Ask more clarifying questions as needed – antigravityguy Mar 4 '14 at 20:25
Is $$A^2+B^2+C^2>=X^2+Y^2$$ – DJohnM Mar 4 '14 at 21:18
Yes. the inequality is true – antigravityguy Mar 4 '14 at 21:45
As, @user58220 points out, there is a vast amount of solutions, I think the question would be vastly more interesting if one more point is fixed. – Sawarnik Mar 5 '14 at 21:14

Is there in fact a unique solution?

Consider a circle, radius $A$, centered at the origin, and another circle, radius $C$, centered at $(X, Y)$ Any line segment, length $B$ with end points anywhere on the two circles, would represent a solution.

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I thought about that. But even if you consider the end of every line segment to be a circle, you can create a right angled triangle by considering the x and y parts to be sin ø and cos ø – antigravityguy Mar 4 '14 at 21:24
See Four-bar linkage... – DJohnM Mar 4 '14 at 21:30
Okay four-bar linkage was very helpful. I'm working on the problem now and will update here when I find solution – antigravityguy Mar 4 '14 at 21:46
This animation, lets you play with parameters and watch all the solutions slide by... – DJohnM Mar 4 '14 at 21:48

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