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Math Help - Ambiguous Case Help

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    Ambiguous Case Help

    Ship Navigation Three ships, A, B, and C, are sailing along the same straight line on a course of 240. Ship A is at the front, and ship C brings up the rear. From ship A. the bearing of a navigation buoy is 025. The buoy is 2.5km from ship A and 2km from ships B and C. To the nearest degree, what is the bearing of the buoy from ship C? from ship B?
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    Re: Ambiguous Case Help

    Quote Originally Posted by Braedong View Post
    Ship Navigation Three ships, A, B, and C, are sailing along the same straight line on a course of 240. Ship A is at the front, and ship C brings up the rear. From ship A. the bearing of a navigation buoy is 025. The buoy is 2.5km from ship A and 2km from ships B and C. To the nearest degree, what is the bearing of the buoy from ship C? from ship B?
    Have you made a sketch of the ships' and buoy's relative positions?
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    Re: Ambiguous Case Help

    Hello, Braedong!

    Three ships, A, B, and C, are sailing along the same straight line on a course of 240.
    Ship A is at the front, and ship C brings up the rear.
    From ship A. the bearing of a navigation buoy is 025.
    The buoy is 2.5km from ship A, and 2km from ships B and C.
    To the nearest degree, what is the bearing of the buoy from ship C? from ship B?

    Code:
          P          Q                  R
    . . . *          *                  *     W
    . . . :          :      X           :     * - - - - - - - - - - - * O
    . . . :          :      o           :                60d    *
    . . . :          :    **   *  2     :                 *
    . . . :          :  * *       *     :           *
    . . . :          :*  *2          *  :      *
    . . . :         *:  *               o
    . . . :       *  : *          *     C
    . . . : 25d *    :*     *
    . . . :   * 5d   o
    . . . : *   *    B
          o
          A
    The ships are at A.B,C. . The buoy is at X. . AX = 2.5,\;BX = CX = 2

    \angle W\!O\!A = 60^o,\quad \angle P\!AO = \angle QBO = \angle RCO = 30^o

    \angle P\!AX = 25^o \quad\Rightarrow\quad \angle X\!AC = 5^o


    In \Delta AXC, Law of Sines: . \frac{\sin(\angle XCA)}{2.5} \,=\,\frac{\sin5^o}{2}

    . . \sin(\angle XCA) \,=\,0.108944678 \quad\Rightarrow\quad \angle XCA \:\approx\:6^o

    \angle RCA = 150^o \quad\Rightarrow\quad \angle RCX = 150^o - 6^o \,=\,144^o

    Therefore, the bearing of the buoy from ship C is 216^o.


    Since \Delta BXC is isosceles, \angle XBC = \angle XCA = 6^o

    Hence: \angle QBX \:=\:30^o - 6^o \:=\:24^o

    Therefore, the bearing of the buoy from ship B is 024^o.

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    Re: Ambiguous Case Help

    how about this sketch? amazing what can be done w/ MS Paint ...
    Attached Thumbnails Attached Thumbnails Ambiguous Case Help-ship-problem.jpg  
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    Re: Ambiguous Case Help

    I probably should have left the answers but Ship C is at 286 and ship B is at 14. That seems to conflict with Soroban's answer though
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    Re: Ambiguous Case Help

    Quote Originally Posted by Braedong View Post
    I probably should have left the answers but Ship C is at 286 and ship B is at 14. That seems to conflict with Soroban's answer though
    I agree w/ those solutions. Understand that my "sketch" is not to scale.
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    Re: Ambiguous Case Help

    I've determined that B(uoy)AC is 30 and angle BCB(uoy) is 102.6 but from there I don't know how to determine the bearings of C in correlation with the buoy. How do you do that?
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    Re: Ambiguous Case Help

    Quote Originally Posted by Braedong View Post
    I've determined that B(uoy)AC is 30 and angle BCB(uoy) is 102.6 but from there I don't know how to determine the bearings of C in correlation with the buoy. How do you do that?
    no ... B(uoy)AC = 35 degrees

    let \theta = angle B(uoy)CA

    \frac{\sin{\theta}}{2.5} = \frac{\sin(35^\circ)}{2}

    then note that angle B(uoy)BA = 180 - \theta
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