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Thread: Word bearing Problem

  1. #1
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    Word bearing Problem

    A ship is sailing due north. At a certain point the bearing of a lighthouse is N $\displaystyle 48.6$E and the distance is 12.5. After a while, the captain notices that the bearing of the lighthouse is now S $\displaystyle 42.9$ E. How far did the ship travel between the two observations of the lighthouse.

    What's the distance?
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  2. #2
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    Hello, qbkr21!

    A ship is sailing due north.
    At a certain point the bearing of a lighthouse is $\displaystyle N\,48.6\,E$ and the distance is 12.5 (miles?)
    After a while, the captain notices that the bearing of the lighthouse is now $\displaystyle S\,42.9\,E$.
    How far did the ship travel between the two observations of the lighthouse?
    Code:
        B *
          | *
          |   *
          |42.9*
          |       *
         c|         *
          |      88.5* C
          |        *
          |48.6* b=12.5
          |  *
        A *

    At $\displaystyle A$ the ship takes a bearing of the lighthouse at $\displaystyle C$
    . . and finds angle $\displaystyle A = 48.6^o$ and $\displaystyle b = AC = 12.5$ miles.

    Later at point $\displaystyle B$, they find that angle $\displaystyle B = 42.9^o$.
    . . Hence, angle $\displaystyle C\:=\:180^o - 48.6^o - 42.9^o \:=\:88.5^o$

    To find side $\displaystyle c$, use the Law of Sines: .$\displaystyle \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}$

    So we have: .$\displaystyle c \;= \;\frac{12.5\sin88.5^o}{\sin42.9^o} \;=\;18.35659392$

    Therefore, the distance is about $\displaystyle 18.4$ miles.

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  3. #3
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    Re:

    Thanks Soroban you are incredible at math. I will learn from this.
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