Hello zacharyrod Originally Posted by
zacharyrod How would I find the simplest radical form of 36 degrees? The attatchment displays the situation.
This is all to do with the Golden Ratio.
In the diagram, let $\displaystyle BC = x$. Then $\displaystyle AD= DC = x$ (since $\displaystyle \triangle$'s $\displaystyle BDC$ and $\displaystyle ADC$ are isosceles). Therefore $\displaystyle BD = 1 - x$.
Now use the fact that the $\displaystyle \triangle$'s $\displaystyle ABC$ and $\displaystyle BDC$ are similar: $\displaystyle \frac{AC}{BC}=\frac{BC}{BD}$
$\displaystyle \Rightarrow \frac{1}{x}=\frac{x}{1-x}$
$\displaystyle \Rightarrow 1-x = x^2$
$\displaystyle \Rightarrow x^2+x-1=0$
$\displaystyle \Rightarrow x = \frac{-1+\sqrt{5}}{2}$, taking the positive root
and now$\displaystyle \cos 36^o = \frac{\tfrac12AC}{DC}$$\displaystyle =\frac{1}{-1+\sqrt5}$
$\displaystyle =\frac{1+\sqrt5}{4}$
Grandad