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Math Help - 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 48.6E and the distance is 12.5. After a while, the captain notices that the bearing of the lighthouse is now S 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 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 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 A the ship takes a bearing of the lighthouse at C
    . . and finds angle A = 48.6^o and b = AC = 12.5 miles.

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

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

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

    Therefore, the distance is about 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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