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Thread: Help Géo !!!

  1. #1
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    Help Géo !!!

    Hii !

    I need Help For This Exercice :

    Either $\displaystyle ABC$ a triangle and $\displaystyle D$ point belonging to $\displaystyle [BC]$ such as :

    $\displaystyle \widehat{DAC} = 90$ , $\displaystyle \widehat{BAD} = 30$ , $\displaystyle BD=AC=1$ .


    Find / Calculate $\displaystyle CD$ .
    Last edited by Perelman; Feb 5th 2010 at 08:30 AM.
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  2. #2
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    Quote Originally Posted by Perelman View Post
    Hii !

    I need Help For This Exercice :

    Either $\displaystyle ABC$ a triangle and $\displaystyle \bold{\color{red}D}$ point <<<<<< typo?
    belonging to $\displaystyle [BC]$ such as :

    $\displaystyle \widehat{DAC} = 90$ , $\displaystyle \widehat{BAD} = 30$ , $\displaystyle BD=AC=1$ .


    Find / Calculate $\displaystyle CD$ .
    This is only a first step and not the final solution.

    The radii have the length 1.
    Attached Thumbnails Attached Thumbnails -cd_in3eck.png  
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  3. #3
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    Quote Originally Posted by Perelman View Post
    Hii !

    I need Help For This Exercice :

    Either $\displaystyle ABC$ a triangle and $\displaystyle A$ point belonging to $\displaystyle [BC]$ such as :

    $\displaystyle \widehat{DAC} = 90$ , $\displaystyle \widehat{BAD} = 30$ , $\displaystyle BD=AC=1$ .


    Find / Calculate $\displaystyle CD$ .
    1. In the right triangle DCS you have:

    $\displaystyle 1+s^2 = x^2$..........[1]

    2. In the triangle ABD you have:

    $\displaystyle s \cdot \sin(30^\circ) = p~\implies~p=\tfrac12 s$

    3. In the right triangle left of A you have:

    $\displaystyle h = 1 \cdot \sin(60^\circ)~\implies~h=\tfrac12 \cdot \sqrt{3}$

    4. Now use similar right triangles:

    $\displaystyle \dfrac ph = \dfrac1{1+x}~\implies~\dfrac{\tfrac12 s}{\tfrac12 \cdot \sqrt{3}}=\dfrac1{1+x}$

    which yields: $\displaystyle s^2 = \dfrac3{(1+x)^2}$ ..........[2]

    5. Combine [1] and [2] and you have to solve for x:

    $\displaystyle 1+ \dfrac3{(1+x)^2} = x^2$

    6. After moving some stuff around you'll get:

    $\displaystyle \tfrac12 x^4 + x^3 - x -2 = 0$

    7. You can show that x = -2 must be a solution of the equation, that means you can factor the LHS:

    $\displaystyle \frac12 \cdot (x+2)(x^3-2)=0$

    8. Therefore the only possible solution is $\displaystyle x = \sqrt[3]{2}~\approx~ 1.259921...$
    Attached Thumbnails Attached Thumbnails -cd_in3eck.png  
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