Computing a Geometrical ratio $\frac{a}{b}$. $XYZ$  is an equilateral triangle as shown on the image below. 
The aim is to find the ratio $\frac{a}{b}$.
So far from the picture, it is easy to see that $b= \frac{YZ}{2}$. 
Does anyone have an idea on how to find $b$ in term of $a$?
It seems from discussion with my classmates that this might lead to another geometrical interpretation of the Golden ratio.

 A: Use the power of the point $R$ (or $Q$) with respect to the big circle.:
$$ (a+b)\cdot a = s^2$$
where $s= YZ/2$. 
Note that since $QRX$ is equlateral we have $s=b$. Now write $q= b/a$ and we get$$ q^2-q-1=0\implies q={1\pm \sqrt{5}\over2}$$
A: 
Another way to find the ratio
is to solve 
the $\triangle ROS$,
for which we know two sides and an angle:
\begin{align} 
\triangle ROS:\quad
|OR|&=r=\tfrac13\cdot2b\cdot\tfrac{\sqrt3}2
=\tfrac{\sqrt3}3\,b
,\\
|OS|&=2r=\tfrac{2\sqrt3}3\,b
,\\
\angle SRO&=\angle ZRO+\angle SRZ=90^\circ+60^\circ
=150^\circ
,
\end{align}
so we can apply the cosine rule to get
\begin{align} 
|OS|^2&=|OR|^2+|RS|^2-2\cdot|OR|\cdot|RS|\cos\angle SRO
,\\
\tfrac43\,b^2&=
\tfrac13\,b^2+a^2-2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos150^\circ
,\\
\tfrac43\,b^2&=
\tfrac13\,b^2+a^2+2\cdot\tfrac{\sqrt3}3\,b\cdot a\cdot\cos30^\circ
,
\end{align}
\begin{align} 
b^2
-b\cdot a
-a^2
&=0
,\\
(\tfrac{b}a)^2
-
\tfrac{b}a
-1
&=0
,
\end{align}
which gives the answer (positive root) as
\begin{align} 
\frac{b}a
&=\frac{1+\sqrt{5}}{2}
,
\end{align}  
which is indeed the Golden ratio.
