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Math Help - Tirg word problems

  1. #1
    Junior Member
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    Tirg word problems

    I have two word problems that I can't solve for the life of me. This one I tried to solve using vectors and the properties and formulas but it didnt work.
    1. Two planes leave an airport at the same time, one is going north at 400 mph (45 degrees) and the other is going directly west at 300 mph. How far apart are they after 2 hours from take off.

    My second problem is similar I just really dont know where to go. I have drawn pictures and I am just lost.

    2. A 300-ft broadcast antenna stands at the tip of a hill whose sides are inclined at 18.6 degrees to the horizontal. How far down the the hill will a 250 ft. support cable extend if it is attached halfway up the antenna?

    Thanks to anyone who can help!
    AC
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, AC!

    The second problem took some manuvering . . .


    2. A 300-ft broadcast antenna stands at the tip of a hill
    whose sides are inclined at 18.6 to the horizontal.
    How far down the the hill will a 250 ft. support cable extend
    if it is attached halfway up the antenna?
    Code:
                                        o B
                                        |
                                        | 150
                                        |
                                        o C
                                      * |
                              250   *   | 150
                                  *     |
                                *       o D
                           F  *   *     :
                            o      x    :
                      *                 :
                *  18.6                :
        A o - - - - - - - - - - - - - - o E

    \Delta DEA is the hill. . \angle A = 18.6^o

    BD is the antenna: . BC = 150,\;CD = 150

    CF is the cable: . CF = 250

    We want: . x = FD


    In right triangle DEA\!:\;\angle A = 18.6^o \quad\Rightarrow\quad \angle ADE = 71.4^o \quad\Rightarrow\quad \angle CDF = 108.6^o



    In \Delta CDF, use the Law of Sines: . \frac{\sin(\angle CFD)}{150} \:=\:\frac{\sin108.6^o}{250}

    . . \sin(\angle CFD) \:=\:0.568661046 \quad\Rightarrow\quad \angle CFD \:\approx\:34.6^o

    Then: . \angle C \:=\:180^o - 108.6^o - 34.6^o \;=\;36.7^o


    In \Delta CDF, we have: . \frac{x}{\sin36.7^o} \:=\:\frac{250}{\sin108.6^o}


    . . Therefore: . x \;=\;157.6400787 \;\approx\;157.6\text{ ft}

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  3. #3
    Junior Member
    Joined
    Jul 2008
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    Wow

    Thanks
    Your awsome make sense now!~
    AC
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