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?

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.
