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Math Help - 2 triangles

  1. #1
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    2 triangles

    THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad

    Prove that triangle ABC and BPC are similar triangles . Hence deduce that

    (1) BC^2=CP.CA

    (2) \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)

    I am only stucked with (2) , i am ok with the rest . THanks .
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  2. #2
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    if your diagram is right, then BC = BP is wrong.
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  3. #3
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    Quote Originally Posted by ukorov View Post
    if your diagram is right, then BC = BP is wrong.
    why is it so
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    Quote Originally Posted by thereddevils View Post
    why is it so
    oh alright not wrong actually, but could have been more detailed and more accurate.
    Last edited by ukorov; November 9th 2009 at 10:29 AM.
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  5. #5
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad

    Prove that triangle ABC and BPC are similar triangles . Hence deduce that

    (1) BC^2=CP.CA

    (2) \cos (\frac{\pi}{5})=\frac{1}{4}(\sqrt{5}+1)

    I am only stucked with (2) , i am ok with the rest . THanks .
    (2) Suppose BP=BC=AP = d. Then in \triangle BPC
    \angle BCP = \tfrac{2\pi}{5}

    \Rightarrow PC = 2d\cos(\angle BCP) = 2d\cos(\tfrac{2\pi}{5})
    So, using the result in (1):
    d^2 = 2d\cos(\tfrac{2\pi}{5})\Big(d+2d\cos(\tfrac{2\pi}{  5})\Big)

    \Rightarrow 1 = 2\cos(\tfrac{2\pi}{5})+4\cos^2(\tfrac{2\pi}{5})
    Solving the quadratic, taking the positive root gives
    \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}

    \Rightarrow 2\cos^2(\tfrac{\pi}{5})-1 = \frac{\sqrt5-1}{4}

    \Rightarrow \cos^2(\tfrac{\pi}{5})= \frac{\sqrt5+3}{8}
    =\frac{5+2\sqrt5+1}{16}

    =\frac{(\sqrt5+1)^2}{16}
    \Rightarrow \cos(\tfrac{\pi}{5}) = \frac{\sqrt5+1}{4}
    Grandad
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  6. #6
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    Quote Originally Posted by Grandad View Post
    Solving the quadratic, taking the positive root gives
    \cos(\tfrac{2\pi}{5})=\frac{\sqrt5-1}{4}

    \Rightarrow 2\cos^2(\tfrac{\pi}{5}) -1 = \frac{\sqrt5-1}{4}
    why \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1 ??
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  7. #7
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    Hello ukorov
    Quote Originally Posted by ukorov View Post
    why \cos(\tfrac{2\pi}{5}) = 2\cos^2(\tfrac{\pi}{5})-1 ??
    Because \cos2\theta = 2\cos^2\theta -1.

    Grandad
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