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Math Help - Trig application question -- Help

  1. #1
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    Trig application question -- Help

    From a position 110km northwest of a coast guard station, an oil tanker makes radio
    contact with the coast guard. The tanker is traveling due south at 25km/h. The radar
    unit at the coast guard station has a range of 90km. For what length of time can the
    coast guard expect the tanker to be visible on the radar screen, to the nearest tenth
    of an hour?
    Answer(Suppose to be, anyway): 3.6h

    Can someone please explain how to do this problem correctly? i worked it out before but i keep getting 4.5h as the answer. I cant think of any alternate ways of tackling it and I keep getting 4.5h, how do you solve this problem?

    any help greatly appricated,

    ~Coukapecker~
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  2. #2
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    Hello, Coukapecker!

    From a position 110 km northwest of a coast guard station,
    an oil tanker makes radio contact with the coast guard.
    The tanker is traveling due south at 25 km/hr.
    The radar unit at the coast guard station has a range of 90 km.
    For what length of time can the coast guard expect the tanker
    to be visible on the radar screen, to the nearest tenth of an hour?
    Answer: 3.6 hrs
    Code:
        T *
          : *
        _ :   *  110
     55√2 :     *
          :       *
          :     45 * 
        B * - - -_- - * S
          :   55√2
          :
    The tanker is at T, sailing toward B at 25 km/hr.
    The station is at S.
    Since \angle S = 45^o, then: . TB\,=\,BS\,=\,\frac{110}{\sqrt{2}} \:=\:55\sqrt{2}



    Code:
        T *
          |
          |
          |
        A *
          |  *
          |     *   90
          |        *
          |           *
          |            θ *
        B * - - - - -_- - - * S
                  55/2
    Some time later, the tanker has moved to point A,
    . . where AS \,=\,90 km.

    Let \theta = \angle ASB

    Then: . \cos\theta \:=\:\frac{55\sqrt{2}}{90} \:=\:0.864241621 \quad\Rightarrow\quad \theta \:\approx\:30.2^o

    And: . \sin30.2^o \,=\,\frac{AB}{90}\quad\Rightarrow\quad AB \:=\:90\sin30.2^o \:\approx\;45.27 km.


    The tanker will be in range from 45.27 km north of B to 45.27 km south of B.

    That is. It will be in range for 90.54 km.

    At 25 km/hr, it will be in range for: . \frac{90.54}{25} \:\approx\:3.6 hours.

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  3. #3
    MHF Contributor
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    Quote Originally Posted by Coukapecker View Post
    From a position 110km northwest of a coast guard station, an oil tanker makes radio
    contact with the coast guard. The tanker is traveling due south at 25km/h. The radar
    unit at the coast guard station has a range of 90km. For what length of time can the
    coast guard expect the tanker to be visible on the radar screen, to the nearest tenth
    of an hour?
    Answer(Suppose to be, anyway): 3.6h

    Can someone please explain how to do this problem correctly? i worked it out before but i keep getting 4.5h as the answer. I cant think of any alternate ways of tackling it and I keep getting 4.5h, how do you solve this problem?

    any help greatly appricated,

    ~Coukapecker~
    Using Pythagoras' theorem, since the tanker is 45 degrees above and to the left of the line of latitude of the station...

    x^2+x^2=110^2

    2x^2=110^2

    x=\frac{110}{\sqrt{2}}

    x^2+y^2=90^2

    y=\sqrt{90^2-\frac{110^2}{2}}=45.28

    2y='90km\ range\ of\ station'=90.56km

    The tanker travels at 25km/hr, so the time interval the tanker remains within range is

    \frac{90.56}{25}=3.62\ hours
    Attached Thumbnails Attached Thumbnails Trig application question -- Help-tanker.jpg  
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