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.
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.
$c = || AB ||;$
$\theta_c = 360 - \theta_a - \theta_b;$
$b = {\sin(\theta_b) \over \sin(\theta_c)} c;$
$C.x = b \cos(\theta_a)$
$C.y = b \sin(\theta_a)$
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 $x,y$
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.
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.