How was this approximation of $\pi$ involving $\sqrt{5}$ arrived at? The Wikipedia article for Approximations of $\pi$ contains this little gem:
$$
\pi \approx \frac{63}{25}\times\frac{17 + 15\sqrt{5}}{7 + 15\sqrt{5}}
$$
which is clearly in $\mathbb{Q[\sqrt{5}]}$.  Wikipedia doesn't (currently) give a reference for this approximation.  I also noticed that when re-written to move $\sqrt{5}$ out of the denominator, the resulting number in $\mathbb{Q[\sqrt{5}]}$ is
$$
\pi \approx \frac{31689 + 4725\sqrt{5}}{13450}
$$
and the integers involved are no larger than $5$ significant digits.  How do you think this approximation was arrived at, and how might one go about finding a better approximation using more digits for the integers?
See also this related question.
 A: This is not yet a complete answer, but may be useful.
The largest root of the polynomial 
$$269x^2-503x+209$$
is 
$$\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$
Changing the polynomial to
$$(25)^2\times269x^2-25\times63\times 503 x+63^2\times 209$$
modifies the root to the approximation given.
$$\pi\approx\frac{63}{25}\times\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$
In terms of the golden ratio, this is
$$\pi\approx\frac{63}{25}\left(1+\dfrac{1}{3\phi-\dfrac{4}{5}}\right)$$
This seems related to the simpler polynomial
$$25x^2-90x+36$$
that can have its coefficients factored as
$$5^2x^2-5\times 6 \times 3x+6^2$$
and has the largest root
$$\pi\approx \frac{9}{5}+\sqrt{\frac{9}{5}},$$
which is another approximation by Ramanujan that has a similar expression in terms of the golden ratio.
$$\pi\approx \dfrac{6}{5}\left(1+\dfrac{1}{\phi-1}\right)$$
This suggests an intermediate approximation of the form
$$\pi \approx r_2\left(1+ \dfrac{1}{2\phi+d_2}\right)$$
and hopefully better precision with an expression similar to
$$\pi \approx r_4\left(1+ \dfrac{1}{4\phi+d_4}\right)$$
or 
$$\pi \approx r_5\left(1+ \dfrac{1}{5\phi-d_5}\right)$$
if even multiples of $\phi$ are of no use. Fractions $r_n$ and $d_n$ are yet to be found.
A: When I first came across your question, I thought it was a modern-day approximation by somebody using a computer. But when d125q pointed out it was by Ramanujan, then I figured out he must have used a systematic method. 
One way is to use a Ramanujan-Sato pi formula like,
$$\frac{1}{\pi} = \frac{1}{16}\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6}\frac{(42\phi-6)n+(5\phi-3)}{(2^{12}\phi^8)^n}\tag1$$
where $\phi=\frac{1+\sqrt{5}}{2}$, and truncate it as for finite number of terms. For example, using just $n=0\;\text{to}\;1$, and getting the reciprocal, it yields,
$$\pi \approx \frac{2^{13}}{3(-383+560\sqrt{5})}$$
It is only good for $10^{-7}$, and the next is $10^{-10}$, but there is an infinite choice of $n$. 
There are three formulas in Mathworld that use a $\sqrt{5}$, including a version of $(1)$. And there is also a fourth. However, Ramanujan must have known still another formula because I can't get the approximation in your post by truncating any of the four.
A: The very general rule with these close approximations is that they're arrived at by finding a continued fraction which has an unusually large term early on. You can then stop at that term and the error is small.
For example, taking the simple continued fraction for $\pi:$
$[3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...]$
If you stop at the $292$ you get an approximation good to $6$ decimal places.

Therefore to find the next very good approximation similar to your example given, find a continued fraction it's taken from, and stop at its next large term.

$\pi$ is related to the golden ratio $\Phi$ by:
$\pi=\frac{5}{\Phi}\cdot\frac{2}{\sqrt{2+\sqrt{2+\Phi}}}\cdot\frac{2}{\sqrt{\sqrt{2+\sqrt{2+\Phi}}}}\cdots$
Which is thanks to John Baez and Greg Egan
I suspect with a bit of jiggery pokery that might yield your example and its successors.
