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Math Help - Trig Question about Bearings

  1. #1
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    Trig Question about Bearings

    Jason is on a 50m high bridge and sees two boats anchored below. From his position, Boat A has a bearing of 230 degrees and boat B has a bearing of 120 degrees. Jason estimates the angles of depression to be 38 degrees for boat A and 35 degrees for Boat B. How far apart are the boats. Nearest meter.

    Im just a bit confused about the bearings part.

    Help is appreciated Thanks in advance
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, ferken!

    This is a 3-D trig problem, but we can work in two dimensions at a time.


    Jason is on a 50m high bridge and sees two boats anchored below.
    From his position, boat A has a bearing of 230 and boat B has a bearing of 120.
    Jason estimates the angles of depression to be 38 for boat A and 35 for Boat B.
    How far apart are the boats (nearest meter)?
    Jason looks at boat A.
    Code:
          W - - - - - * J
                38 * |
                  *   |
                *     | 50
              *       |
            * 38     |
        A * - - - - - * C
    Jason is at J.
    The bridge is: JC = 50
    \angle WJA = 38^o = \angle JAC

    In right triangle JCA\!:\;\;\tan38^o \:=\:\frac{50}{AC} \quad\Rightarrow\quad AC \:=\:\frac{50}{\tan38^o} \:\approx\:64.0



    Jason looks at boat B.
    Code:
        J * - - - - E
          | * 35
          |   *
       50 |     *
          |       *
          |     35 *
        C * - - - - - * B
    Jason is at J.
    The bridge is: JC = 50
    \angle EJB = 35^o = \angle JBC

    In right triangle JCB\!:\;\;\tan35^o \:=\:\frac{50}{BC} \quad\Rightarrow\quad BC \:=\:\frac{50}{\tan35^o} \:\approx\:71.4



    Looking down from the sky, this is the diagram:
    Code:
                    N
                    |
                    |
                    |
                   C* 120
               64 *   *
                *  110 *  71.4
             A*           *
                    *       *
                          *   *
                                *B

    Law of Cosines: . AB^2 \;=\;64^2 + 71.4^2 - 2(64)(71.4)\cos110^o \;=\;12319.75049

    Therefore: . AB \;=\;110.9943715 \;\approx\;111\text{ m}

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