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Math Help - Hot Air Balloon Angle with Ground

  1. #1
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    Apr 2007
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    Hot Air Balloon Angle with Ground

    We had this question to do for homework which I am very confused with, can anyone help me?

    Jennifer and Alex were flying a hot air balloon when they decided to calcutlate the straight line distance one the ground from Mathville to Trigtown. From a vertical height of 340m, they measured a 2degree angle of depression to Mathville, and a 3degree andlge of depression to Trigtown. The angle between these two sight lines, from the balloon, to the two towns was 80degrees. Find the distance along the ground from Mathville to Trigtown. Round to the nearest tenth of a metre.

    It also asks for a 3-dimensional well labeled sketch, I dont understand how you would draw this in 3-D.
    Sorry its such a big question but I really need help. Thank you so much in advance
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Sarah!

    Jennifer and Alex were flying a hot air balloon when they decided to calculate
    the straight line distance on the ground from Mathville to Trigtown.
    From a vertical height of 340m, they measured a 2° angle of depression to Mathville,
    and a 3° angle of depression to Trigtown.
    The angle between these two sight lines, from the balloon, to the two towns was 80°.
    Find the distance along the ground from Mathville to Trigtown to the nearest tenth of a metre.
    I won't even try to type a 3-D diagram for this problem . . .


    Consider the triangle involving the balloon and Mathville.
    Code:
          B
          * - - - - - - - - -
          |   *  2°
          | 88°   *
      340 |           *
          |               *
          |                   *
          * - - - - - - - - - - - *
          G           a           M
    The balloon is at B directly above point G on the ground; Mathville is at M.
    The angle of depression is 2°, hence: / GBM = 88°.
    . . And we have: .tan 88° = a/340 . . a .= .340·tan88° . .9736.326


    Consider the triangle involving the balloon and Trigtown.
    Code:
                                  B
                  - - - - - - - - *
                          3°  *   |
                          *   87° |
                      *           | 340
                  *               |
              *                   |
          * - - - - - - - - - - - *
          T           b           G
    The balloon is at B direcly above point G on the ground; Trigtown is at T.
    The angle of depression is 3°, hence: /TBG = 87°.
    . . And we have: .tan 87° = b/340 . . b .= .340·tan87° . .6487.586


    Looking down at the ground, we have this triangle:
    Code:
                    M
                    *
                   *  *
         9736.326 *     *
                 *        *
                *           *
               * 80°          *
            G *  *  *  *  *  *  * T
                    6487.586

    We can apply the Law of Cosines to determine side MT.

    . . MT˛ .= .9736.326˛ + 6487.586˛ - 2(9736.326)(6487.586)cos 80°

    I'll let you crank it out . . .

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  3. #3
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    Joined
    Apr 2007
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    Thanks you soooo much for your help, it means alot. seriously it does. lol
    this 3-d model is still confusing me....its probably the simplest part of the whole thing. Hmmmm *thinks some more*
    But once again thank you!
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