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 cos36^° cos72^° .

I filled in the remaining angles and lengths, showing triangles PCB and PCD to be isosceles triangles with angles of 108^°, 36^° and 36^° and sides PC = PB = PD = 1 unit.

fig1 I found cos36^° and cos72 ^° by using the cosine rule for triangles DCP and APD respectively.

cos36^° = x² + 1 - 1
2x
= x
2

cos72^° = 1 +x² - x²
2x
= 1
2x

Combining these two expressions,

cos36^° cos72^° = x
2
. 1
2x
= 1
4
.

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

1
2
x² sin108^° = 1
2
x·1·sin72^° + 1
2
1·1·sin108^°.

As sin108^° = sin72^°, this gives x²=x+1 and hence x is the golden ratio:

x = 1+Ö5
2
= j.

Hence

cos36^° - cos72^° = x
2
- 1
2x
= x²-1
2x
= 1
2
.

fig 2 Let the common side of the two triangles be x. Then we have: b = x·cos72^° and
a = x
cos36^°
.

Therefore

b
a
= cos72^° cos36^° = 1
4

so a is four times bigger than b.

Here we have
x + b
a
=cos36^°

and ^x/_a = cos72^° so

b
a
= cos36^° -cos72^° = 1
2
.

Therefore a is twice the length of b. fig 3