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I have a triangle ABC in the x-y coordinate plane.

Corners A and B lie on x-axis. Length of AB is known. Angles A and B are also known.

Question: Find (x,y) coordinates of corner C.

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3  
You know the three angles, then you can use the law of sines for find the other sides –  Bryan Yocks Dec 14 '10 at 23:39

4 Answers 4

Another possibility is as follows: this solution to your problem involves using the fact that the slope of a line making an angle $\theta$ with the positive x-axis is $\tan\;\theta$; now, what you have are two lines with slopes $\tan\;A$ and $\tan(\pi-B)$ in one case, and slopes $\tan\;B$ and $\tan(\pi-A)$ in the other. Construct the point-slope form for the two lines through $A$ and $B$, and find their intersection through the usual methods.

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Hint:

With law of sines you can find the other sides.

If $A= (x_{1},y_{1})$ and $B=(x_{2},y_{2})$, then $|AB|=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$

Now, if $C=(x,y)$ get :

$$|AC| = \sqrt{(x-x_{1})^2+(y-y_{1})^2}$$ $$|BC| = \sqrt{(x-x_{2})^2+(y-y_{2})^2}$$

Remember that you know $|AB|,|AC|,|AB|$.

Now you can find $x,y$

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A system of linear equations:

Assume A is at the origin.

You can form basis vectors for AC and BC using the angles A and B

U_{AC} = [cosA  sinA]^T
U_{BC} = [cosB  sinB]^T

Then we know geometrically

w*U_{AC} - x*U_{BC} = [ AB  0 ]^T

This is a 2x2 linear system. Solve for w and x.

Then

x = w*cosA
Y = w*sinA

best regards.

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up vote 0 down vote accepted

$c = || AB ||;$

$\theta_c = 360 - \theta_a - \theta_b;$

$b = {\sin(\theta_b) \over \sin(\theta_c)} c;$

$C.x = b \cos(\theta_a)$

$C.y = b \sin(\theta_a)$

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This is assuming that A is at the origin –  Mikhail Dec 15 '10 at 19:37

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