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Math Help - Bearing problem jd

  1. #1
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    Apr 2009
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    Bearing problem jd

    This is a question i have been stuck on for hours. please help!


    The following is all that is given in the question:

    An airplane leaves an airport on a bearing of 45degrees.
    After flying for 65km, the airplane turns and flies on a bearing of 240 degrees for 62 km. To the nearest km, how far is the airplane fom the airport?
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  2. #2
    Super Member

    Joined
    May 2006
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    Lexington, MA (USA)
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    Hello, giles84!

    I assume you understand "bearings."


    An airplane leaves an airport on a bearing of 45.
    After flying for 65 km, the airplane turns and flies on a bearing of 240 for 62 km.
    To the nearest km, how far is the airplane fom the airport?

    It leaves the airport A on a bearing of 45 and flies 65 km to B.
    Code:
                                        o B
                                      * :
                                    *45:
                                  *     :
                                *
                              *
                            *
                :         * 65
                :       *
                :     *
                :45*
                : *
                o
                A

    Then it flies on a heading of 240 for 62 km to P.
    Code:
                                        :
                                   120 :
                                        o B
                                   *     240
                      62      *
                         *
                    *
               *
        P o

    The final diagram looks like this:
    Code:
                                        :
                                   120 :
                                        o B
                                   *  * :
                      62      *     *   :
                         *   15  * 45 :
                    *           *       :
               *              *
        P o                 *
           *              * 65
            *   :       *
             *  :     *
              * :45*
               *: *
                o
                A

    We find that \angle PBA = 15^o
    And we can use the Law of Cosines.

    PA^2 \;=\;65^2 + 62^2 - 2(65)(62)\cos15^o \;=\;283.6378401


    Therefore: . PA \;=\;16.841552 \;\approx\;17 km.

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