Find the coordinates in an isosceles triangle Given:
$A = (0,0)$
$B = (0,-10)$
$AB = AC$
Using the angle between $AB$ and $AC$, how are the coordinates at C calculated?

Let $a,b$ and $c$ be the side lengths and $A,B$ and $C$ the angles.
$a^{2}=x^{2}+\left( y+10\right) ^{2}$
$b^{2}=x^{2}+y^{2}=10^{2}$
$b=c=10$
By the (Neper) theorem of tangents (corollary of the Law of tangents):
$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}$
On the other hand
$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $
and by the theorem of sinus
$c\sin A=a\sin C\iff \left( x^{2}+\left( y+10\right) ^{2}\right) \sin
C=10\sin A$
Compiling, we get:
$\frac{A-B}{2}=\arctan (\frac{\sqrt{x^{2}+\left( y+10\right) ^{2}}-10}{\sqrt{%
x^{2}+\left( y+10\right) ^{2}}+10}\cot \frac{C}{2})$
$\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}\quad C<\pi $
$(x^{2}+(\sqrt{100-x^{2}}+10)^{2})\sin C=10\sin A$
$y^{2}=10^{2}-x^{2}$
We have to solve this system of four equations and four unknowns $x,y,A,B$.
Edit: I started this approach before the question has been updated.
