# Math Help - Cos 36 degrees to simplest radical form?!

1. ## Cos 36 degrees to simplest radical form?!

How would I find the simplest radical form of 36 degrees? The attatchment displays the situation.
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2. 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 $BC = x$. Then $AD= DC = x$ (since $\triangle$'s $BDC$ and $ADC$ are isosceles). Therefore $BD = 1 - x$.

Now use the fact that the $\triangle$'s $ABC$ and $BDC$ are similar:
$\frac{AC}{BC}=\frac{BC}{BD}$

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

$\Rightarrow 1-x = x^2$

$\Rightarrow x^2+x-1=0$

$\Rightarrow x = \frac{-1+\sqrt{5}}{2}$, taking the positive root

and now
$\cos 36^o = \frac{\tfrac12AC}{DC}$
$=\frac{1}{-1+\sqrt5}$

$=\frac{1+\sqrt5}{4}$
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