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Math Help - [SOLVED] Trig Bearing Problems

  1. #1
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    Exclamation [SOLVED] Trig Bearing Problems

    #1 Starting at point A, a ship sails 18.5 km on a bearing of 189, then turns and sails 47.8 km on a bearing of 317. Find the distance of the ship from point A.

    I'd asked for help on yahoo answers, and the people there told me this

    x = 18.5cos 189 + 47.8cos 317 = 16.686
    y = 18.5sin 189 + 47.8sin 317 = -35.494
    distance = (16.686) + (-35.494) = 39.220 km

    it's the correct answer in the book, but I dont understand where the equations came from..



    #2 What bearing and airspeed are required for a plane to fly due north in 2.5 hours if the wind is blowing from a direction of 328 at 11 mph?

    I know that the plane is currently at 160 mph now, but then I dont know how the picture would even look like so I'm not sure where to begin...



    Please help, I hate bearing problems
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  2. #2
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    Hello, gofish!

    I think it's easier with the Law of Cosines.


    1) Starting at point A, a ship sails 18.5 km on a bearing of 189,
    then turns and sails 47.8 km on a bearing of 317.
    Find the distance of the ship from point A.
    Code:
                          N
        C *         M     :
           \  *     :     :
            \     * :     :
             \      : *   :
              \     :     * A
          47.8 \    :    /:
                \43:   / :
                 \  :9/9:
                  \ : /   :
                   \:/    :
                    *     :
                    B     S

    The ship sails from A to B\!:\;\;AB = 18.5
    Hence, major angle NAB \,= \,189^o\quad\Rightarrow\quad \angle SAB = \angle ABM = 9^o

    Then it sails from B\text{ to }C\!:\;\;BC = 47.8
    Major angle MBC = 317^o\quad\Rightarrow\quad \angle MBC = 43^o
    . . Hence: . \angle ABC = 52^o

    Law of Cosines: . AC^2 \;=\;47.8^2 + 18.5^2 - 2(47.8)(18.5)\cos52^o \;=\; 1538.231115


    Therefore: . A \:=\:39.22028959 \;\approx\;39.220 km.

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  3. #3
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    ohhh I see I see. I drew the picture in a different, more weird way. that's probably why I got confused. Thanks a ton

    I still need help on #2 though....
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  4. #4
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    Quote Originally Posted by gofish View Post
    #2 What bearing and airspeed are required for a plane to fly due north in 2.5 hours if the wind is blowing from a direction of 328 at 11 mph?

    I know that the plane is currently at 160 mph now, but then I dont know how the picture would even look like so I'm not sure where to begin...
    how far north in 2.5 hrs?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    how far north in 2.5 hrs?
    ahh I forgot to put the distance! sorry

    400 miles due north in 2.5 hrs
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  6. #6
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    the triangle is a rough sketch of the navigation triangle for this problem.

    AB is the wind vector ... it has a length of 11 mph

    CB is the ground vector ... it has a length of 160 mph

    angle B = 148 degrees

    using the law of cosines, the air vector, CA = 169.4 mph

    now use the law of sines to calculate angle C.
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  7. #7
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    so that's how it's drawn! I didn't know where the wind was supposed be at.. thanks!! I understand it now.
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