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Math Help - Bearing Problems

  1. #1
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    Bearing Problems

    For some reason I am having a terrible time with some bearing problems that were assigned. Any help would be greatly appreciated.
    1. An airplane flew 450mi at a bearing of N65*E from Airport A to Airport B. The plane then flew at a bearing of S38*E to airport C. Find the distance from A to C if the bearing from airport A to Airport C is S60*E *=degrees
    I can't get my angles to work out I know to use the law of sines, but I'm getting the wrong angles?

    2. The Sides of a parallelogram are 10 feet and 14 feet. The longer diagonal is 18 feet. Find the length of the shorter diagonal of the parallelogram.
    I don't understand how the shorter diagonal would look? |\| but wouldn't the shorter diagonal just make it look like an x inside of a box?

    3. The heading and air speed of an airplane are 60* and 250 mph, respectively. If the wind is 50 mph with a direction of 150*, find the ground speed and resulting course (direction) of the plane.
    I have no idea where to start this problem?
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  2. #2
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    Quote Originally Posted by jasonk View Post
    For some reason I am having a terrible time with some bearing problems that were assigned. Any help would be greatly appreciated.
    1. An airplane flew 450mi at a bearing of N65*E from Airport A to Airport B. The plane then flew at a bearing of S38*E to airport C. Find the distance from A to C if the bearing from airport A to Airport C is S60*E *=degrees
    I can't get my angles to work out I know to use the law of sines, but I'm getting the wrong angles?

    ...
    Make a sketch. Calculate the angles and then use Sine rule.
    Attached Thumbnails Attached Thumbnails Bearing Problems-flugzeug_dreieck.png  
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  3. #3
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    Quote Originally Posted by jasonk View Post
    ...

    2. The Sides of a parallelogram are 10 feet and 14 feet. The longer diagonal is 18 feet. Find the length of the shorter diagonal of the parallelogram.
    I don't understand how the shorter diagonal would look? |\| but wouldn't the shorter diagonal just make it look like an x inside of a box?

    ...
    1. Use the Cosine rule to calculate the magnitude of the angle \phi

    2. Since |\theta| = 180^\circ - |\phi| you can use the Cosine rule to calculate the length of the smaller diagonal.
    Attached Thumbnails Attached Thumbnails Bearing Problems-strecken_parallelogramm.png  
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  4. #4
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    Hello, jasonk!

    You can start by making a sketch . . .


    3. The heading and air speed of an airplane are 60 and 250 mph, respectively.
    The wind is 50 mph with a direction of 150.
    Find the ground speed and resulting course (direction) of the plane.
    Code:
          P                 Q
          :                 :
          :                 : 150
          :               B o
          :               * : * 50
          :             *60:30*
          :      250  *     :     o C
          :         *       : *
          :       *       * :
          :     *     *     :
          :60*   *         R
          : * *
        A o

    The plane's heading is: \angle PAB = 60^o,\;AB = 250

    The wind's heading is: \angle QBC = 150^o,\;BC = 50


    Note that \angle ABC = 90^o

    Then AC is the hypotenuse of right triangle ABC.
    . . AC \:=\:\sqrt{250^2 + 50^2} \:=\:\sqrt{65,\!000} \;\approx\;255


    In right triangle ABC\!:\;\;\tan(\angle BAC) \:=\:\frac{50}{250}\:=\:0.2

    . . then: . \angle BAC \:=\:\arctan(0.2) \;\approx\;11.3^o

    Hence: . \angle PAC \:=\:60^o + 11.3^o \:=\:71.3^o


    Therefore, the plane's heading is 71.3 and its speed is 255 mph.

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