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

How would I find the simplest radical form of 36 degrees? The attatchment displays the situation.

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 $\displaystyle BC = x$. 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
$\displaystyle \cos 36^o = \frac{\tfrac12AC}{DC}$
$\displaystyle =\frac{1}{-1+\sqrt5}$

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

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### cos 36 degrees

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