How would I solve for the bases of an isosceles without the height? In the image below if you were to take 36 degrees and set it between 0-180 degrees how would you solve for the bases between points AC. There are special cases for the bases length such as (60 degrees where the length is 5) and (180 degrees where the length is 10). However a problem arises when (90 degrees where the length is 7.07) that when doubled is (180 degrees where the length is 14.14). Why is that? 
Isosceles triangle with no base or height
I have also looked online but have found no articles or solutions to this specific problem. I want to have a better understanding so what do you suggest?
 A: You can use trigonometry to determine the base. Denote the midpoint created by the altitude of the triangle point $D$.
Since triangle $ABD$ is right, we may apply our trigonometric formulae. Specifically, we know that $\sin 18^{\circ}=\frac{AD}{5}$. I found $18^{\circ}$ because the altitude bisects the top angle of the triangle. 
From this relationship, can you solve for $AD$? Then, can you solve for the total length of the base, knowing that $AD$ is one half of $AC$? 
EDIT: If you want to express $\sin 18^{\circ}$ as radicals, it can be shown that $\sin 18^{\circ}=\frac{1}{4}(\sqrt{5}-1).$
A: A $36-72-72$ triangle is a special case involving the golden ratio.
That is, the ratio of base to height is $1:\frac{1+\sqrt5}{2}$.
Therefore, the base length is the following:
$$\frac{10}{1+\sqrt 5}$$
A: Without trig:

You know $|AB|=|BC|=5$ and seek $x=|AC|$. (From there, the height is easy to find using the Pythagorean theorem.)
Extend $AC$ and place $D$ and $E$ on it, so that $\measuredangle BDA = \measuredangle CEB = 36^\circ$. Notice, now, that the triangles $\triangle BAD$ and $\triangle DBE$ are similar, as they both have angles $36^\circ,36^\circ,108^\circ$.
This means that the ratios $\frac{|BD|}{|DE|}$ and $\frac{|AB|}{|BD|}$ are equal.
Notice, also, that $\triangle BCD$ is isosceles, so $|BD|=|CD|=|AC|+|AB|=x+5$.
The triangles $\triangle BAD$ and $\triangle BEC$ are isosceles, so $|DA|=|AB|=5$ and $|CE|=|BC|=5$. Thus, $|DE|=|DA|+|AC|+|CE|=x+10$.
So we have: $\frac{x+5}{x+10} = \frac{5}{x+5}$ or $x^2+5x-25=0$.
The only positive root of this quadratic equation is $x=\frac{5 \sqrt 5 - 5}2$.
