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Math Help - Trigonometry Question with bearings

  1. #1
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    Trigonometry Question with bearings

    I need help on text book question: (I am confused, I think that enough information is not given) please help me if you know how solve the problem.

    Question:
    A submarine is being followed by two ships, A and B, 3.8 km apart, with A due east of B. If A is on a bearing of 165 degree from the submarine and B is on the bearing of 205 degree from the submarine, find the distance from the submarine to both ships.

    Answer given at the back of text book is: 5.4km and 5.7km
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  2. #2
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    Measure "bearing from the submarine" from the direction the submarine is travelling and with reference to a viewer on the submarine..

    Bearing 0 from the submarine would be directly in front.

    Bearing 180 from the submarine would be directly in behind.

    Bearing 90 from the submarine would be directly to the starboard.

    Bearing 270 from the submarine would be directly to the port.
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  3. #3
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    Hello, lalji!

    We must understand "bearings" and make a good sketch . . .


    A submarine is being followed by two ships, A\text{ and }B,
    3.8 km apart, with A due east of B.

    If A is on a bearing of 165 from the submarine
    and B is on the bearing of 205 from the submarine,
    find the distance from the submarine to both ships.

    Answers: 5.4 km and 5.7 km
    Code:
                      N
                      :
                      : 90
                     S* - - - E 
                    *  * 75
                  * 40 *
              a *        * b
              *           *
            * 65      75 *
        B *  *  *  *  *  *  * A
                  3.8

    We have: . AB = 3.8

    Since \angle NSA = 165^o,\text{ then: }\,\angle ESA = \angle A = 75^o
    Since major \angle NSB = 205^o\!:\;\angle ASB = 40^o

    In \Delta BSA\!:\;\angle B \:=\:180^o - 40^o - 75^o \:=\:65^o
    Let  a = SB,\;b = SA


    Law of Sines: . \frac{a}{\sin75^o} \:=\:\frac{3.8}{\sin40^o} \quad\Rightarrow\quad a \:=\:5.710312527

    Law of Sines: . \frac{b}{\sin65^o} \:=\:\frac{3.8}{sin40^o}\quad\Rightarrow\quad b \:=\:5.357865551


    Therefore: . \boxed{\begin{array}{ccc}\text{A-to-S}&\approx &  5.4\text{ km} \\ \text{B-to-S}&\approx&5.7\text{ km} \end{array}}

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  4. #4
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    Thanks for the response

    You have explained it very well. Thank you very much.
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