<s>[INST] 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.
 [/INST] We have $A(45,10),B(10,20),C(x_c,y_c)$.
$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$
$C=B=20^{%
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}=\pi /9$ rad
$A=180^{%
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}-40^{%
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}=140^{%
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}=7\pi /9$ 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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