How to get the third point coordinates in isosceles triangle?

Question

Isosceles triangle $ABC$
$AB = AC = d_1$
$BC = d_2$

$A = (x_1, y_1)$ $B = (x_2, y_2)$ $C = (x_3, y_3)$

$\angle BAC = \phi$
$\angle ABC =\angle ACB = \theta$

I want an equation for $x_3$ and $y_3$ (and I know there will be two values)

Answer 1

I could solve this. The solution key was the conversion between the Cartesian Coordinate System and the Polar Coordinate System.

$x_3 = AB * Cos(ϕ) + x_1$

$y_3 = AB * Sin(ϕ) + y_1$

Answer 2

If you have all the sides, you don't need the angles-which do you believe? Taking the sides, you have two equations in two unknowns: one for the distance from $A$ and one from the distance from $B$. So $(x_3-x_1)^2+(y_3-y_1)^2=d_1^2$ and $(x_3-x_2)^2+(y_3-y_2)^2=d_2^2$ You are correct there will be two solutions. With two quadratics you might expect four, but two will be extraneous.