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?
 A: edit (to match revised question): Given your revised question, there is still the issue of C being on either side of the y-axis, but you have specified that AB=AC and that you are given $\mathrm{m}\angle BAC$ (the angle between AB and AC), so as in my original answer (below), the directed (trigonometric) measure of the angle from the positive x-axis to AC is $\mathrm{m}\angle BAC-90^{\circ}$ and AC=AB=10, so C has coordinates $(10\cos(\mathrm{m}\angle BAC-90^{\circ}),10\sin(\mathrm{m}\angle BAC-90^{\circ}))$.  (This matches up to one of the answers in Moron's solution; the other corresponds to the other side of the y-axis.)
original answer (when it was not specified that AB=AC and when the given angle was C):  As suggested in the comments, there are several cases.  First, C could be on either side of the y-axis; let's assume that C has positive x-coordinate (leaving the case where it has negative x-coordinate for you to solve).
Second, ABC could be isosceles with AB=AC, AB=BC, or AC=BC.  In the first case, $\angle B\cong \angle C$ (which cannot happen unless C is acute) and $\mathrm{m}\angle BAC=180^{\circ}-2\mathrm{m}\angle C$, so the directed (trigonometric) measure of the angle from the positive x-axis to AC is $90^{\circ}-2\mathrm{m}\angle C$ and AC=AB=10, so C has coordinates $(10\cos(90^{\circ}-2\mathrm{m}\angle C),10\sin(90^{\circ}-2\mathrm{m}\angle C))$.  The second case is similar to the first (so it's left for you to solve).  In the third case, C is equidistant from A and B, so C must lie on the perpendicular bisector of AB (as in J. Mangaldan's comment), and by symmetry this perpendicular bisector of AB also bisects $\angle ACB$; from there, you can use right triangle trigonometry to determine the coordinates of C (left for you to solve).
The cases where AB=AC (blue), AB=BC (red), and AC=BC (green) (lighter versions on the left side of the y-axis) are shown below for measures of angle C between 0 and 180°.

A: Use Polar Co-ordinates. A point with polar co-ordinates $(r,\theta)$ is the same point in $x,y$ co-ordinates (or as also called, rectangular co-ordinates) as $(r\cos \theta, r\sin \theta)$.
In this case, point $C$ lies at a distance $10$ from $A$ which is the origin, so $r = 10$.
If given angle CAB is $\alpha$, the the polar angle is either $\frac{3pi}{2}-\alpha$  or $\frac{3pi}{2}+\alpha$, i.e in polar co-ordinates $C$ is either $(10,\frac{3pi}{2}-\alpha)$ or $(10,\frac{3pi}{2}+\alpha)$. 
(It might help to draw a figure).
Now convert back to $x,y$ co-ordinates.
A: 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.
