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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)

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  • $\begingroup$ Hint $$y_2=y_3$$ $$x_1=\frac{x_2+x_3}{2}$$ $\endgroup$ – Shabbeh Jul 28 '14 at 2:59
  • $\begingroup$ @Vincent I searched and found similar questions but couldn't find the exact one nor the wanted answer, I've been trying for about 4 hours but couldn't find it yet.. I tried using Pythagorean theorem, law of sines and law of cosines... $\endgroup$ – Aly Elhaddad Jul 28 '14 at 3:04
  • $\begingroup$ I don’t quite understand what you are asking. The following are my guesses. (1) $d_1$, $d_2$, ϕ, and θ are the givens; (2) want to find $x_3$ (also $y_3$) in terms of the givens. If they are true, I have the following comments:- [1] either ϕ or θ is known will be sufficient because one is derivable from the other. [2] d1 and d2 are also inter-derivable with the help of θ (or ϕ). [3] Do a series of transformations, to reduce your triangle to the one being symmetrically located about the y-axis and the midpoint of BC at the origin. Then $y_3 = 0$, and $x_3 = - x_2 = d_1 cos θ$. $\endgroup$ – Mick Jul 28 '14 at 5:39
  • $\begingroup$ @Mick , You guessed right. About your third comment, can you clarify? $\endgroup$ – Aly Elhaddad Jul 28 '14 at 17:57
  • $\begingroup$ @AlyEl-Haddad Transformations include translate [or commonly called shift] (left/right/up/down); rotate (through certain angles); reflect (about an axis). These transformations are shape and size preserving. Any object, after going through a combination of these transformations, can be “reduced” to a more convenient location. In this way, the original problem can be reduced to a much more simplified equivalent (like those I have suggested). If the reduced version can be solved, the original problem is then considered as solved also. $\endgroup$ – Mick Jul 29 '14 at 1:51
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I could solve this. The solution key was the conversion between the Cartesian Coordinate System and the Polar Coordinate System.

$x_3 = AB * Cos(ϕ) + x_1$

$y_3 = AB * Sin(ϕ) + y_1$

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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.

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  • $\begingroup$ thanks but I could find this already find this I need something like x_3$ = ... y_3$ = ... $\endgroup$ – Aly Elhaddad Jul 28 '14 at 3:19
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Since AC = $d_1$,
$d_1^{2}$ = $(x_3-x_1)^2$ + $(y_3-y_1)^2$ $\ \ \ \ \ \ \ \ $ _$1$

And since BC = $d_2$
$d_2^{2}$ = $(x_3-x_2)^2$ + $(y_3-y_2)^2$ $\ \ \ \ \ \ \ \ $ _$2$

Now there are two unknowns and two equations.
Using these two equations, we can get $x3$ and $y3$.

$y_3 = y_1 \pm\sqrt{d_1^2-(x_3-x_1)^2}$

Substituting this value of $y_3$ from $eqn_1$ into $eqn_2$, we can get $x_3$.

$x_3 = \dfrac{-4 d_1^2 x_1 + 4 d_1^2 x_2 \pm \sqrt{(4 d_1^2 x_1 - 4 d_1^2 x_2 - 4 d_2^2 x_1 + 4 d_2^2 x_2 - 4 x_1^3 + 4 x_1^2 x_2 + 4 x_1 x_2^2 - 4 x_1 y_1^2 + 8 x_1 y_1 y_2 - 4 x_1 y_2^2 - 4 x_2^3 - 4 x_2 y_1^2 + 8 x_2 y_1 y_2 - 4 x_2 y_2^2)^2 - 4 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2) (d_1^4 - 2 d_1^2 d_2^2 - 2 d_1^2 x_1^2 + 2 d_1^2 x_2^2 - 2 d_1^2 y_1^2 + 4 d_1^2 y_1 y_2 - 2 d_1^2 y_2^2 + d_2^4 + 2 d_2^2 x_1^2 - 2 d_2^2 x_2^2 - 2 d_2^2 y_1^2 + 4 d_2^2 y_1 y_2 - 2 d_2^2 y_2^2 + x_1^4 - 2 x_1^2 x_2^2 + 2 x_1^2 y_1^2 - 4 x_1^2 y_1 y_2 + 2 x_1^2 y_2^2 + x_2^4 + 2 x_2^2 y_1^2 - 4 x_2^2 y_1 y_2 + 2 x_2^2 y_2^2 + y_1^4 - 4 y_1^3 y_2 + 6 y_1^2 y_2^2 - 4 y_1 y_2^3 + y_2^4)} + 4 d_2^2 x_1 - 4 d_2^2 x_2 + 4 x_1^3 - 4 x_1^2 x_2 - 4 x_1 x_2^2 + 4 x_1 y_1^2 - 8 x_1 y_1 y_2 + 4 x_1 y_2^2 + 4 x_2^3 + 4 x_2 y_1^2 - 8 x_2 y_1 y_2 + 4 x_2 y_2^2}{2 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2)}$

Similarly,
$x_3 = x_1 \pm\sqrt{d_1^2-(y_3-y_1)^2}$

Substituting this value of $x_3$ from $eqn_1$ into $eqn_2$, we can get $y_3$.

$y_3 = \dfrac{-4 d_1^2 y_1 + 4 d_1^2 y_2 \pm \sqrt{(4 d_1^2 y_1 - 4 d_1^2 y_2 - 4 d_2^2 y_1 + 4 d_2^2 y_2 - 4 x_1^2 y_1 - 4 x_1^2 y_2 + 8 x_1 x_2 y_1 + 8 x_1 x_2 y_2 - 4 x_2^2 y_1 - 4 x_2^2 y_2 - 4 y_1^3 + 4 y_1^2 y_2 + 4 y_1 y_2^2 - 4 y_2^3)^2 - 4 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2) (d_1^4 - 2 d_1^2 d_2^2 - 2 d_1^2 x_1^2 + 4 d_1^2 x_1 x_2 - 2 d_1^2 x_2^2 - 2 d_1^2 y_1^2 + 2 d_1^2 y_2^2 + d_2^4 - 2 d_2^2 x_1^2 + 4 d_2^2 x_1 x_2 - 2 d_2^2 x_2^2 + 2 d_2^2 y_1^2 - 2 d_2^2 y_2^2 + x_1^4 - 4 x_1^3 x_2 + 6 x_1^2 x_2^2 + 2 x_1^2 y_1^2 + 2 x_1^2 y_2^2 - 4 x_1 x_2^3 - 4 x_1 x_2 y_1^2 - 4 x_1 x_2 y_2^2 + x_2^4 + 2 x_2^2 y_1^2 + 2 x_2^2 y_2^2 + y_1^4 - 2 y_1^2 y_2^2 + y_2^4)} + 4 d_2^2 y_1 - 4 d_2^2 y_2 + 4 x_1^2 y_1 + 4 x_1^2 y_2 - 8 x_1 x_2 y_1 - 8 x_1 x_2 y_2 + 4 x_2^2 y_1 + 4 x_2^2 y_2 + 4 y_1^3 - 4 y_1^2 y_2 - 4 y_1 y_2^2 + 4 y_2^3}{2 (4 x_1^2 - 8 x_1 x_2 + 4 x_2^2 + 4 y_1^2 - 8 y_1 y_2 + 4 y_2^2)}$

Links to solutions

$x3$:
http://www.wolframalpha.com/input/?i=solve+d2%5E2%3D(x3-x2)%5E2+%2B+(sqrt(d1%5E2+-+(x3+-+x1)%5E2)+%2B+y1+-+y2)%5E2+for+x3

$y3$:
http://www.wolframalpha.com/input/?i=solve+d2%5E2%3D(sqrt(d1%5E2+-+(y3+-+y1)%5E2)+%2B+x1+-x2)%5E2+%2B+(y3+-+y2)%5E2+for+y3

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