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Thread: Geometry Problem

  1. #1

    Question Geometry Problem

    The Vriend Monument is a sacred place for Geometry students in Western Albania. Chance Jones uses a surveying device (set on ground) to measure the angle of elevation to the top of the monument to be 3.2 degrees. He then walks one mile (5,280 feet) closer and measures the angle of elevation to be 15.4 degrees. How high (in feet) is the monument?

    If someone can could you give me a full explanation? Thank you in advanced.

  2. #2

    Geometry problem

    I am having trouble solving for the variables on c and d on this website. PLease help I'm confused

    http://oh-but-its-cold-outside.tumblr.com/

  3. #3
    Elite Member
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    Hello, Ernesto223!

    The Vriend Monument is a sacred place for Geometry students in Western Albania.
    Chance Jones measures the angle of elevation to the top of the monument to be 3.2 o.
    He then walks one mile (5,280 feet) closer and the angle of elevation is 15.4 o.
    How high (in feet) is the monument?

    Code:
        V *
          | * *
          |   *   *
        h |     *     *
          |        *       *
          |         *         *
          |     15.4o *      3.2o *
        M *-------------*-------------* A
          : - -  x  - - B - - 5280  - :
    The height of the monument is: [tex]h = VM.[/tex]

    Jones stands at [tex]A[/tex] and [tex]\angle VAM = 3.2^o.[/tex]
    He moves to [tex]B\!:\;AB = 5280\text{ ft},\;\angle VBM = 15.4^o.[/tex]
    Let [tex]x = MB.[/tex]

    In [tex]\Delta VMA\!:\;\tan3.2^o \:=\:\dfrac{h}{x+5280} \quad\Rightarrow\quad x \:=\:\dfrac{h-5280\tan3.2^o}{\tan3.2^o}[/tex] .[1]

    In [tex]\Delta VMB\!:\;\tan15.4^o \:=\:\dfrac{h}{x} \quad\Rightarrow\quad x \:=\:\dfrac{h}{\tan15.4^o}[/tex] .[2]

    Equate [1] and [2]: .[tex]\dfrac{h-528\tan32^o}{\tan3.2^o} \:=\:\dfrac{h}{\tan15.4^o} [/tex]

    . . [tex]h\tan15.4^o - 5280\tan3.2^o\tan15.4^o \:=\:h\tan3.2^o[/tex]

    . . [tex]h\tan15.4^o - h\tan3.2^o \;=\;5280\tan3.2^o\tan15.4^o[/tex]

    . . [tex]h(\tan15.4^o - \tan3.2^o) \;=\;5280\tan3.2^o\tan15.4^o[/tex]

    . . [tex]h \;=\;\dfrac{5280\tan3.2^o\tan15.4^o}{\tan15.4^o - \tan3.2^o} \;=\;370.3747019[/tex]


    Therefore: .[tex]h\;\approx\;370.4\text{ ft}[/tex]

  4. #4
    Elite Member
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    Quote Originally Posted by brnnpink4 View Post
    I am having trouble solving for the variables on c and d on this website. PLease help I'm confused

    http://oh-but-its-cold-outside.tumblr.com/
    Please start your own thread.
    I'm just an imagination of your figment !

  5. #5
    Full Member
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    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

  6. #6
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    Wink Thanks

    Hay thanks i really needed that but how would i draw that picture that would help me a lot.

  7. #7
    Elite Member
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    CoCoD ?
    Are you using various names?

    To "draw that picture", use a pencil and a ruler...
    I'm just an imagination of your figment !

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