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Math Help - Law of Sines....I think

  1. #1
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    Law of Sines....I think

    Two lookout stations, which are 25 miles apart along the coast on a north-south line, spot an approaching yacht. One lookout station measures the direction to the yacht at N33* E, and the other station measures the direction of the yacht at S62* E. How far is the yacht from each lookout station? How far is the yacht from the coast?

    I have no idea where to start for this one. Please help me!! Thanks!
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  2. #2
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    Hello, suzdisp!

    Did you make a sketch?


    Two lookout stations, which are 25 miles apart along a north-south costline,
    spot an approaching yacht.
    One lookout station finds the direction to the yacht at N33E,
    and the other station finds the direction of the yacht at S62E.

    (1) How far is the yacht from each lookout station?

    (2) How far is the yacht from the coast?
    Code:
      -  B*
      :   |  *
      :   | 62 *   a
      :   |        *
      :   |           *
      :   |              *
      :  D* - - - - - - 85 *C
      :   |               *
     25   |             *
      :   |           *
      :   |         *
      :   |       * b
      :   |     *
      :   |33*
      :   | *
      -  A*

    Since \angle A = 33^o,\:\angle B = 62^o, then: . \angle C \:=\:180^o - 33^o - 62^o \:=\:85^o


    Law of Sines: . \frac{a}{\sin A} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad a \:=\:\frac{c\sin A}{\sin C}

    . . a \:=\:\frac{25\sin33^o}{\sin86^o} \;=\;13.66798666 \quad\Rightarrow\quad\boxed{ a \;\approx\;13.7\text{ miles}} (1)


    Law of Sines: . \frac{b}{\sin B} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad b \:=\:\frac{c\sin B}{\sin C}

    . . b \:=\:\frac{25\sin62^o}{\sin85^o} \:=\:22.15800773 \quad\Rightarrow\quad\boxed{ b \;\approx\;22.2\text{ miles}} (1)


    In right triangle CDA\!:\;\;\sin33^o \:=\:\frac{CD}{b}

    Therefore: . CD \:=\:22.2\sin33^o \:=\:12.09068658\quad\Rightarrow\quad\boxed{ CD \:\approx\:12.1\text{ miles}} (2)

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