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I have two points and all sides of right triangle

I need find A point

\begin{gather*} |AB| = 1 \\ |BC| = 1 \\ |AC| = \sqrt{1^2 + 1^2} = \sqrt2 \\ A(?,?) \\ B(0,0) \\ C(1,0) \\ \\ |AB| = 1 \\ |BC| = \sqrt{1^2 + 1^2} = \sqrt2 \\ |AC| = \sqrt{1^2 + \sqrt2^2} = \sqrt3 \\ A(?,?) \\ B(0,0) \\ C(1,0) \end{gather*}

The angle in $B$ is 90 degrees.

I need a formula to find point $A$. i need infinity loop that create a chain of triangle not just one triangle

MHA
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  • $|BC|=\sqrt2$ and $B(0,0),C(1,0)$ are in contradiction since $(1,0)$ and $(0,0)$ are evidently 1 apart, not $\sqrt2$ apart. If I follow the picture, the answer to the first set of equations is terribly obvious: $A(0,1)$. And all you need is the positions of $B,C$, the angle $A\hat BC=90$ and $|AB|=1$, without the other lengths. Of course, $A(0,-1)$ would be equally fine with just sides and angles, but then the picture rules $A(0,-1)$ out, right? – MickG Dec 18 '15 at 21:23
  • @fleablood this was more or less my reaction too :). – MickG Dec 18 '15 at 21:37
  • I'm sorry. It was rude. – fleablood Dec 18 '15 at 21:38
  • I will now add the square roots I twice forgot to add :). – MickG Dec 18 '15 at 21:39
  • The suggestion to make a drawing was a good point though. Perhaps you should have removed the "LOOK" part and left the rest of the comment :). – MickG Dec 18 '15 at 21:42

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