This is probably not well phrased but I cannot think exactly how to phrase it. If I have a triangle such that the first point is on fixed axis (0,0) and the second point is free to move, where would I find the third point please? The triangle is an isosceles and this point is located at the end of the two matching sides - the 'elbow' if you like. I appreciate that it can be in one of two places so is probably linked to a square rule, but cannot think which.
Many thanks in advance.
If your triangle has base $b$ and height $h$, then you have one point at $(0,0)$, one at $(b\cos(\alpha),b\sin(\alpha))$ for some angle $\alpha$, and the third point at $(\frac{b\cos(\alpha)}{2}\mp h \sin(\alpha),\frac{b\sin(\alpha)}{2}\pm h \cos(\alpha))$
Alternatively, if you know the base $b$ and the hypotenuse $H$, you have the internal angle $\beta$ at the base $$H\cos(\beta)=b/2$$ $$\beta=\arccos\left(\frac{b}{2H}\right)$$ and then you have your points at $(b\cos(\alpha),b\sin(\alpha))$ and $(H\cos(\alpha\pm\beta),H\sin(\alpha\pm\beta))$
