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. ${\displaystyle \cos {36}^{\circ }=\frac{x^2+1-1}{2x}=\frac{x}{2}}$ ${\displaystyle \cos {72}^{\circ }=\frac{1+x^2-x^2}{2x}=\frac{1}{2x}}$ Combining these two expressions, ${\displaystyle \cos {36}^{\circ }\cos {72}^{\circ }=\frac{x}{2}.\frac{1}{2x}=\frac{1}{4}.}$

Consider the area of the triangle `above' the diagonal, and express it is the sum of two areas :

${\displaystyle \frac{1}{2}{x}^2\sin {108}^{\circ }=\frac{1}{2}x·1·\sin {72}^{\circ }+\frac{1}{2}1·1·\sin {108}^{\circ }.}$

As $\sin {108}^{\circ }=\sin {72}^{\circ }$, this gives $x^2=x+1$ and hence $x$ is the golden ratio:

${\displaystyle x=\frac{1+\sqrt{5}}{2}=\phi .}$

Hence

${\displaystyle \cos {36}^{\circ }-\cos {72}^{\circ }=\frac{x}{2}-\frac{1}{2x}=\frac{x^2-1}{2x}=\frac{1}{2}.}$

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

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