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Math Help - 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
    Grandad's Avatar
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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 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}

    Grandad
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