Using Degrees Where A intercepts C the Y coordinate is Phi The X coordinate is (sin^-1(1/ phi^2))/ 2
Where B intercepts C the Y coordinate is 1/Phi The X coordinate is (sin^-1(1/phi) )/ 2
I got these equations while messing around with Area-Diagonal relationships.
Basically replaced x and y with sin and cos.
Why does the golden ratio show up here?
Interesting observation. We need the identities $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$.
To solve $A$ and $C$, we set
$$\begin{align}(\sin x + \cos x)^2&=\frac {2 \sin x \cos x}{(\sin x - \cos x)^2} \\(\sin^2 x + 2\sin x\cos x+\cos^2 x)(\sin^2 x - 2\sin x\cos x+\cos^2 x)&=2 \sin x \cos x\\(1+\sin 2x)(1 - \sin 2x)&=\sin 2x \\1-\sin^2 2x&=\sin 2x \\\sin^2 2x+\sin 2x-1&=0\end{align}$$
and this equation leads to the golden ratio, since $$\sin 2x = \frac {-1\pm\sqrt5}2 \implies (\sin x+\cos x)^2=1+\sin2x=\frac {1\pm\sqrt5}2$$ Similarly for $B$ and $C$:
$$\begin{align}(\sin x - \cos x)^2&=\frac {2 \sin x \cos x}{(\sin x - \cos x)^2} \\(\sin^2 x - 2\sin x\cos x+\cos^2 x)^2&=2 \sin x \cos x\\(1 - \sin 2x)^2&=\sin 2x \\1-2\sin 2x + \sin^2 2x&=\sin 2x \\\sin^2 2x-3\sin 2x+1&=0\end{align}$$
and we thus have
$$\sin 2x = \frac {3\pm\sqrt5}2 \implies (\sin x-\cos x)^2=1-\sin2x=\frac {-1\pm\sqrt5}2$$
