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Thread: 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 N33°E,
    and the other station finds the direction of the yacht at S62°E.

    (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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