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Math Help - Navigation Problem

  1. #1
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    Navigation Problem

    A ship leaves port at 10 miles per hour, with a heading of N 35 W. There is a warning buoy located 5 miles directly north of the port. What is the bearing of the warning buoy as seen from the ship after 7.5 hours?

    Now I have drawn a diagram and solved that the angle should be 52.7 degrees I just can't figure out what direction. is it N 52.7 W
    or would it be N 52.7 E
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Re: Math problem me and friends are trying to solve

    Hello, pierce!

    I believe your diagram is wrong . . .


    A ship leaves port at 10 mph with a heading of N 35 W.
    There is a warning buoy located 5 miles directly north of the port.
    What is the bearing of the warning buoy as seen from the ship after 7.5 hours?

    Now I have drawn a diagram and solved that the angle should be 52.7 degrees.
    . . Exactly where is that angle?
    I just can't figure out what direction.
    is it N 52.7 W or would it be N 52.7 E?
    How can be the port be to the upper-left or upper-right?

    After 10 hours, the ship has traveled 75 miles.
    The diagram should look like this:


    Code:
          X
          *
          :
          :
          :               Y
        S *               *
          : * *           :
          :   *   *       :
          :     *     *   :
          *    75 *       * B
          Z         *     |
                      *35d|5
                        * |
                          *
                          P

    The port is at P.
    The ship is at S.
    The buoy is at B.
    \angle BPS = 35^o,\;SP = 75

    Law of Cosines:
    . . SB^2 \;=\;5^2 + 75^2 - 2(5)(75)\cos35^o \;=\;5035.635967

    . . . SB \;=\;70.96221506 \;\approx\;71\text{ miles}


    Law of Sines:

    . . \frac{\sin(SBP)}{75} \:=\:\frac{\sin(SPB)}{71} \quad\Rightarrow\quad \sin(SBP) \:=\:\frac{75\sin(SPB)}{71} \:=\:0.605890602

    . . . . \angle SBP \;=\;37.29295617 \;=\;37.3^o\text{ or }142.7^o

    We see that \angle SBP is obtuse: . \angle SBP \,=\,142.7^o

    Then: . \angle XSB \,=\,142.7^o \quad\Rightarrow\quad \angle BSZ \,=\,37.3^o


    Therefore, the heading of the buoy from the ship is: . S\,37.3^o\,E

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