Results 1 to 3 of 3

Math Help - Help Géo !!!

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    46

    Help Géo !!!

    Hii !

    I need Help For This Exercice :

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

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


    Find / Calculate CD .
    Last edited by Perelman; February 5th 2010 at 08:30 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote Originally Posted by Perelman View Post
    Hii !

    I need Help For This Exercice :

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

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


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

    The radii have the length 1.
    Attached Thumbnails Attached Thumbnails Help Géo !!!-cd_in3eck.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote Originally Posted by Perelman View Post
    Hii !

    I need Help For This Exercice :

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

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


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

    1+s^2 = x^2..........[1]

    2. In the triangle ABD you have:

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

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

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

    4. Now use similar right triangles:

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

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

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

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

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

    \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:

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

    8. Therefore the only possible solution is x = \sqrt[3]{2}~\approx~ 1.259921...
    Attached Thumbnails Attached Thumbnails Help Géo !!!-cd_in3eck.png  
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum