Here is some information about regular pentagons.

$ABCDE$ is a regular pentagon.

We first prove that triangles $AEX$ and $ADC$ are similar and that the ratio $AD / AE$ is equal to the golden ratio $\frac{1 + \sqrt{5}}{2}$ .

ABCDE

is a regular pentagon, triangles

AEX

and

ADC

have angles

36^{o}, 72^{o}, 72^{o}

and so they are similar isosceles triangles. Label the side length of the pentagon

s

AE = AX = CD = CX = s

and the chord length

c

AD = EC = c

and

EX = c - s

. From the similar triangles

\frac{c - s}{s} = \frac{s}{c}

so writing

AD / AE = c / s = x

gives

x - 1 = \frac{1}{x}

which is the quadratic equation

x^{2} - x - 1 = 0

and hence, as it must be positive, the ratio

x = AD / AE

is the golden ratio

\frac{1 + \sqrt{5}}{2}

Now we can find the exact values of $\cos 72^{o}$ and $\cos 36^{o}$ .

Drawing a perpendicular from $E$ to $AD$ gives a right angled triangle from which

$\cos 36^{o} = \frac{c}{2 s} = \frac{\sqrt{5} + 1}{4} .$

Drawing a perpendicular from $A$ to $DC$ gives a right angled triangle from which

$\cos 72^{o} = \frac{s}{2 c} = \frac{1}{2 ϕ} = \frac{\sqrt{5} - 1}{4} .$

In explaining the construction of a regular pentagon coordinates are useful.

If the pentagon is inscribed in a unit circle with $A$ the point $(0, 1)$ you can find the exact coordinates of $B$ and $C$ using the exact values of $\cos 72^{o}$ and $\cos 36^{o}$ which we have just found.

If the pentagon is inscribed in a unit circle with $A$ the point $(0, 1)$ then $B = (\sin 72^{o}, \cos 72^{o})$ and $C = (\sin 36^{o}, - \cos 36^{o})$ .

Hint: When you explain the construction focus on the $' y'$ coordinates of $B, C, D$ and $E$ .