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Math Help - Trigonometry help please.....

  1. #1
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    Trigonometry help please.....

    Four rods, AB 20CM long, BC 22.5CM long, CD 25CM long and DA 27.5 long,
    are pin-jointed at ABC and D. If the angle ABC is 120, find

    a. the distance AC
    b. the angle ADC


    thankyou
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  2. #2
    Member
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    May 2008
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    a)

    Using cosine rule...

    AC^2 = 20^2 + 22.5^2 - 2*20*22.5Cos120 = 1356.25

    AC = 36.8 cm

    b)

    Again using cosine rule...

    36.8^2 = 27.5^2 + 25^2 - 2*27.5*25CosADC

    2*27.5*25CosADC = 27.5^2 + 25^2 - 36.8^2

    ADC = Cos^-1((27.5^2 + 25^2 - 36.8^2)/(2*27.5*25)) = 90.0 degrees
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, sa87uk!

    Four rods: AB 20cm long, BC 22.5cm long, CD 25cm long, and DA 27.5 cm long,
    are jointed at A, B, C, D.

    If the angle ABC is 120, find

    a) the distance AC
    b) the angle ADC
    Code:
                     22.5
              B o  *  *  *  o C
               * 120    *    *
              *       *         *
          20 *     *              * 25
            *   *                   *
           * *                        *
        A o  *  *  *  *  *  *  *  *  *  o D
                       27.5

    Law of Cosines: . AC^2 \;=\;AB^2 + BC^2 - 2(AB)(BC)\cos B

    . . AC^2 \;=\;20^2 + 22.5^2 - 2(20)(22.5)\cos120^o \;=\;1356.25

    (a) Therefore: . AC \;\;=36.82729966 \;\approx\;36.8 cm



    Law of Cosines: . \cos(\angle ADC) \;=\;\frac{AD^2 + CD^2 - AC^2}{2(AD)(CD)}

    . \cos(\angle ADC) \;=\;\frac{27.5^2 + 25^2 - 1356.25}{2(27.5)(25)} \;=\;\frac{1}{55}

    (b) Therefore: . \angle ADC \;=\;88.95820115^o \;\approx\;89^o



    Edit: Too fast for me, Sean!
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