# Find the coordinates in an isosceles triangle if the triangle it self is in positive axis

A at $(45,10)$, B at $(10,20)$, $AB=AC$ and angle $C=20$ degree find the coordinates of $C$.suggest the formula so i can write code in Perl.

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We have $A(45,10),B(10,20),C(x_c,y_c)$.

$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$

$C=B=20^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=\pi /9$ rad

$A=180^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }-40^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=140^{% %TCIMACRO{\U{ba}}% %BeginExpansion {{}^o}% %EndExpansion }=7\pi /9$ rad

Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.

The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as

$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$

and the vector $\overrightarrow{AC}$ as

$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan \frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{% 2}{7}\right) )$

Therefore in the original referencial $xy$, we have

$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +45\approx 78.239$

$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi \right) +10\approx 24.837$

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Now, the angle at C is determined by the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$. These vectors can be found by subtracting the coordinates of C from A and B (respectively). $(CA)(CB)\cos C = \overrightarrow{CA}\cdot\overrightarrow{CB}$, and you know the distances CA = AC = AB and the measure of C, and you can compute CB and the dot product in terms of (x,y). This equation guarantees that the measure of C is 20°.