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Math Help - Bearings Help

  1. #1
    Newbie firemonkey's Avatar
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    Bearings Help

    Hi i have a scanned copy of what i need help with. if you could post and email me, that would be AWSOME! but i do need this soon!


    Well i just found out, the file is too big, so, if you could email me, i will send you a copy of the sheet. thnx.

    but yes, i am going to type the problem, but the Page is more help....


    Just had a break through! i just croped the scan to the diagram only, so here is what it says....




    A ship is travelling on a straight course at a constant speed, during fog, to avoid detection.

    The ship's journey is to be logged by two government run observatory posts on a stretch of shoreline running north/south. (see diagram [this is were you might need the diagram.]). The observatories are 10Km apart.



    Task One


    At midnight observatory B detects a sighting of a ship bearing 045(degrees). At the same time observatory T detects the ship at a bearing of 129(degrees)

    You must draw a scaled diagram to represent the situation and calculate the psotion of the ship from both observatory B and T at midnight.



    Email : fire_monkey4me@hotmail.com



    thnx!

    YES I KNOW the image is huge, that is not my fault. sorry![/
    Attached Thumbnails Attached Thumbnails Bearings Help-img_0001.jpg  
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  2. #2
    Super Member

    Joined
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    From
    Lexington, MA (USA)
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    Hello, firemonkey!

    A ship is travelling on a straight course at a constant speed.
    Two observation posts B and T are 10 miles apart on a north-south shoreline.
    T is north of B.

    B detects the ship at bearing 045.
    At the same time T detects the ship at bearing of 129.

    Draw a diagram to represent the situation.
    Calculate the postion of the ship from both B and T.
    Code:
          N
          |
          |
        T o 129
          | *
          |51*
          |     * y
          |       *
          |         *
       10 |       84 o S
          |         *
          |       *
          |     * x
          |45*
          | *
        B o
    The ship is at S.

    \angle TBS = 45^o,\;\angle NTS = 129^o\quad\Rightarrow\quad\angle STB = 51^o

    \angle S \:=\:180^o - 51^o - 45^o \:=\:84^o

    We want: . x \,=\,SB,\;y \,=\,ST


    Law of Sines: . \frac{x}{\sin51^o} \:=\:\frac{10}{\sin84^o}\quad\Rightarrow\quad x \:=\:\frac{10\sin51^o}{\sin84^o} \:=\:7.814266987

    . . Therefore, the ship is about 7.81 miles from B.


    Law of Sines: . \frac{y}{\sin45^o} \:=\:\frac{10}{\sin84^o}\quad\Rightarrow\quad y \:=\:\frac{10\sin45^o}{\sin84^o} \:=\:7.11001723

    . . Therefore, the ship is about 7.11 miles from T.

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