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Math Help - [SOLVED] Triangles and bearing

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    [SOLVED] Triangles and bearing

    A ship rounds a headland by sailing first four nautical miles on a course of 069 then five nautical miles on a course of 295.
    Calculate the distance and bearing of its new position from its original position.
    Last edited by tishtash7; February 6th 2009 at 01:18 AM.
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  2. #2
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    Quote Originally Posted by tishtash7 View Post
    A ship rounds a headland by sailing first four nautical miles on a course of 069 then five nautical miles on a course of 295.
    Calculate the distance and bearing of its new position from its original position.
    Bearings are drawn clockwise with respect to North. See attachment.

    1. Calculate the value of the green angle. (295 - 180 - 69)

    2. Use the Cosine rule to calculate the distance d. (Thick black line)

    3. Use Sine rule to calculate the final bearing (red sector).
    Attached Thumbnails Attached Thumbnails [SOLVED] Triangles and bearing-rund_kap.png  
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  3. #3
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    Hello, tishtash7!

    Did you make a sketch?


    A ship rounds a headland by sailing first four nautical miles on a course of 069
    then five nautical miles on a course of 295.
    Calculate the distance and bearing of its new position from its original position.
    Code:
        C *                   Q
           *  *               :
            *     *           :
             *        *    65:
              *     P     *   :
               *    :     46 * B
                *   :       * :
                 *  :     *   :
                  * :69*  69:
                   *: *       :
                    *         :
                    A         R

    The ship sails 4 miles from A to B.
    . . AB \,=\,4,\;\angle PAB \,=\, 69^o \,=\,\angle ABR

    Then it sails 5 miles from B to C.
    . . BC \,=\,5,\;\angle QBC \:=\:360^o - 295^o \:=\:65^o

    Hence: . \angle CBA \:=\:180^o - 69^o - 65^o \:=\:46^o


    In \Delta ABC use the Law of Cosinse:

    . . AC^2 \:=\:4^2 + 5^2 - 2(4)(5)\cos46^o \:=\:13.21366518

    Therefore:. AC \:\approx\:\boxed{3.6\text{ miles}}


    Law of Cosines again: . \cos(\angle CAB) \:=\:\frac{3.6^2 + 4^2-5^2}{2(3.6)(4)} \:=\:0.1375
    . . Hence: . \angle CAB \:\approx\:82^o

    Then: . \angle CAP \:=\:82^o - 69^o \:=\:13^o

    Therefore, the bearing is: . 360^o - 13^o \:=\:\boxed{347^o}

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