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Math Help - Sin and cos rules work , help please

  1. #1
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    Sin and cos rules work , help please

    hi.
    At midnight a ship sailing due north passes two lightships, A and B, which are 10km apart in a line due east from the ship. Lighship A is closer to the ship than B. At 2am the bearings of the lighship are 149degrees and 142degrees.

    I have draw a diagram to help the problem, and i have tried a few ways of finding the answers but i am soo lost , can you please help i need to;
    find the distance of the ship fromA at 2am, and also find the speed of the ship - the speed i dont know where to begin

    thank you
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  2. #2
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    Hello, Chez!

    At midnight a ship sailing due north passes two lightships, A and B,
    which are 10km apart in a line due east from the ship.
    Lighship A is closer to the ship than B.
    At 2 am the bearings of the lighship are 149 and 142.

    (a) Find the distance of the ship from A at 2am.
    (b) Find the speed of the ship.
    Code:
          N
          |
          |
        Q *
          | * *
          |31* 7*
          |     *     *
        y |       * x     *
          |         *         *
          |           *       52 *
        P * - - - - - - * - - - - - - *
                        A     10      B

    At midnight, the ship is at P.
    Two hours later, it is at Q.
    . . \angle NQA = 149^o\quad\Rightarrow\quad \angle AQP = 31^o
    . . \angle PQB = 142^o\quad\Rightarrow\quad \angle BQA = 7^o
    Let x = QA,\;y = QP


    In right triangle QPA,\;\angle AQP = 31^o\quad\Rightarrow\quad \angle QAP = 59^o \quad\Rightarrow\quad \angle QAB = 121^o


    In triangle QAB,\;\angle B \:=\:180^o - 7^o - 121^o \quad\Rightarrow\quad \angle B\:=\:52^o


    Law of Sines: . \frac{x}{\sin52^o} \:=\:\frac{10}{\sin7^o}\quad\Rightarrow\quad x \:=\:\frac{10\sin52^o}{\sin7^o} \:=\:64.66029369

    (a) Therefore: . x \:\approx\:64.7\text{ km}



    In right triangle QPA,\;\cos31^o\:=\:\frac{y}{x}\quad\Rightarrow\qua  d y \:=\:64.7\cos31^o \:=\:55.45872436

    Hence: .  y \:\approx\:55.5\text{ km}

    (b) Therefore, the ship's speed was: . \frac{55.5\text{ km}}{2\text{ hrs}} \;=\;27.75\text{ km/hr}

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  3. #3
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    Hi, thank for that its great but i dont understand were the 121degrees has come from?
    thnkz
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  4. #4
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    its ok lol iv worked it out

    thnks
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