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Thread: Cos 36 degrees to simplest radical form?!

  1. #1
    Newbie zacharyrod's Avatar
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    Exclamation 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. #2
    MHF Contributor
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    Hello zacharyrod
    Quote Originally Posted by zacharyrod View Post
    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
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