Really easy to guess!

The answer is the golden ratio.In a regular pentagone ABCDE, assume the side length is 1 and take the point F on the diagonal AC so that BCF is isosceles in F. The angles ACB, FCB and FBC are all equal, and equal to pi/5. It follows that CFB = 3pi/5 and ABF = AFB = 2pi/5. Therefore ACF is isosceles in A and AC = AF + FC = 1 + FC. But CFB and ABC are similar triangle. Hence FC/BC = AB/AC that is FC = 1/AC. Finally AC = 1+ 1/AC or

x^2 = x + 1

where x = AC then solving this you get golden ratio

1. Consider the triangle formed by two consecutive sides of a regular pentagon and a diagonal. Let the measure of each of the two consecutive sides be 1 unit.
2. Every interior angle of a regular pentagon measures 108 degrees.

3. So, with these assumptions, the triangle is an ASA Triangle.

4. The measure on a triangle's interior angles is 180, so the measure of the two remaining unknown angles is $180 - 108 = 72.$

5. Since the triangle is isosceles, these angles are congruent, so the measure of each of the unknown angles is $\frac {72}{2}$ = 36.

Now, apply Law of Sines--

$\frac{1}{\sin36} = \frac{x}{\sin108}$

$\frac {\sin108}{\sin36}=x$

x=1.618

Since the regular pentagon's side measures 1, the ratio will be 1.618... or the golden ratio.

Let us have a pentagon $ABCDE$ . Draw segment $\overline{BE}$ . Now, we have a cyclic quadrilateral, $BCDE$ . Letting the length of a diagonal be $x$ , we get $x + 1 = x^2 \rightarrow x^2-x-1=0$ , and use the quadratic formula to obtain $x = \dfrac{1 \pm \sqrt{5}}{2}$ . Since the answer must be positive, our solution is $\dfrac{1 + \sqrt{5}}{2}$ , the Golden Ratio .

We know angle between any two adjacent sides of a regular pentagon is 108°. So, supposing side as x (say) and diagonal as y(say), and applying cosine rule, we get y^2 = 2*(1-cos108°) x^2; which upon solving, we get y/x = 1.6180