The first part of this solution came from
Dorothy, S4, Madras College, St Andrew's and the second part from
Vassil, Y11, Lawnswood High School, Leeds.
Consider the rhombus as illustrated, where $x$ is an unknown
length. We have to find the value of $\cos 36^\circ \cos 72^\circ
.$
I filled in the remaining angles and lengths, showing triangles
$PCB$ and $PCD$ to be isosceles triangles with angles of
$108^\circ$, $36^\circ$ and $36^\circ$ and sides $PC = PB = PD = 1$
unit.

I found $\cos 36^\circ$ and $\cos 72 ^\circ$ by using the
cosine rule for triangles DCP and APD respectively. \[ \cos
36^\circ = {{x^2 + 1  1}\over 2x} = {x\over 2} \] \[ \cos 72^\circ
= {{1 +x^2  x^2}\over 2x} = {1\over 2x} \] Combining these two
expressions, \[\cos 36^\circ \cos 72^\circ = {x\over 2}.{1\over
2x}={1\over 4}.\] 
Consider the area of the triangle `above' the diagonal, and
express it is the sum of two areas : \[{1\over 2} x^2 \sin
108^\circ = {1\over 2} x\cdot1\cdot\sin 72^\circ + {1\over 2}
1\cdot 1\cdot\sin108^\circ.\] As $\sin 108^\circ = \sin 72^\circ$,
this gives $x^2=x+1$ and hence $x$ is the golden ratio: \[x =
{1+\sqrt{5}\over 2} = \varphi .\] Hence \[\cos 36^\circ  \cos
72^\circ = {x\over 2}  {1\over 2x} = {x^21\over 2x}={1\over
2}.\]

Let the common side of the two triangles be $x$. Then we have:
$b = x\cdot \cos 72^\circ$ and $ a= {x\over{\cos 36^\circ}}.$
Therefore \[{b\over a} = \cos72^\circ \cos 36^\circ = {1\over 4}\]
so $a$ is four times bigger than $b$. 


Here we have ${{x + b}\over a} =\cos 36^\circ$ and ${x\over a}
= \cos 72^\circ$ so \[{b\over a} = \cos 36^\circ \cos 72^\circ =
{1\over 2}.\] Therefore $a$ is twice the length of $b$. 
