Triangle in Star

Please note this solution is incomplete - I have no way to prove that this is the largest triangle.

After much playing around with this, the largest triangle I found was in the following configuration:

We can work out the edge length of the triangle $s=AB$ by using the sine rule twice.

In $\Delta APQ$ :

$\begin{aligned} PQ &= \varphi \\ \angle APQ &= 36^{\circ} \\ \angle QAP &= 120^{\circ} \\ \angle PQA &= 24^{\circ} \end{aligned}$

where $\varphi=\frac{1+\sqrt5}{2}$ is the golden ratio.

The sine rule gives $AQ=\frac{\varphi\sin{36}}{\sin{120}}$

In $\Delta BQR$ :

$\begin{aligned} QR &= \varphi \\ \angle BQR &= 180-\angle PQA - \angle PQR = 180-24-108=48^{\circ} \\ \angle QRB &= 36^{\circ} \\ \angle RBQ &= 96^{\circ} \end{aligned}$

So $QB=\frac{\varphi \sin{36}}{sin{96}}$

We then have $s=AB=AQ+QB=2.054\ldots$

The trig functions involved can actually be expressed in radicals; this comes out as

$\begin{aligned} s&=\frac{-5+\sqrt5+5\sqrt{\frac{2}{3}\left(5+\sqrt5 \right)}}{4} \end{aligned}$

Either way, the answer is $\lfloor 1000s \rfloor=\boxed{2054}$ .

However, as I mentioned above, this does not prove that the triangle is the largest possible (OK, the answer was accepted as correct, but that doesn't really count as a proof!). The only ideas I had were to use numerical methods, but I'm not sure any of these would easily give the required accuracy. Can anyone prove it?