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Math Help - Trig Problem

  1. #1
    Junior Member VDestinV's Avatar
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    Exclamation Trig Problem

    A satellite dish that is 5 m high sits on top of a building. From a point at the base of the building, the angles of elevation of the botom and top of the satellite dish are 39.1 degrees and 44.7 degrees. Determine the height of the building, to one decimal place.

    I'm completely lost.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by VDestinV View Post
    A satellite dish that is 5 m high sits on top of a building. From a point at the base of the building, the angles of elevation of the botom and top of the satellite dish are 39.1 degrees and 44.7 degrees. Determine the height of the building, to one decimal place.

    I'm completely lost.
    Always draw a diagram. Here is my "artistic rendition" of the situtation

    Trig Problem-capture.jpg


    We can now use the tangent function to find h by setting up a system of equations

    \tan(39.1^\circ)=\frac{h}{x} and \tan(44.7^\circ)=\frac{h+6}{x}

    From here just solve the system for h.

    Good luck.
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  3. #3
    Junior Member VDestinV's Avatar
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    Quote Originally Posted by We can now use the tangent function to find h by setting up a system of equations

    [tex
    \tan(39.1^\circ)=\frac{h}{x}[/tex] and \tan(44.7^\circ)=\frac{h+6}{x}

    From here just solve the system for h.
    but how do I figure out the x then?
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  4. #4
    Super Member

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    Hello, VDestinV!

    Did you make a sketch?


    A satellite dish that is 5 m high sits on top of a building.
    From a point at the base of the building, the angles of elevation of the botom and top
    of the satellite dish are 39.1 and 44.7.
    Determine the height of the building, to one decimal place.
    Code:
        A *
          | *
        5 |   *
          |     *
        B *       *
          |   *     *
        h |       *   *
          |     39.1 * *
        C * - - - - - - - * D
                  x
    AB is the satellite dish: . AB = 5
    BC is the building: . BC = h
    \angle BDC = 39.1^o,\;\;\angle ADC = 44.7^o
    Let CD = x

    In \Delta BCD\!:\;\tan39.1^o \:=\: \frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan39.1^o}\;\;{\color{blue}[1]}

    In \Delta ACD\!:\;\;\tan44.7^o \:=\:\frac{h+5}{x}\quad\Rightarrow\quad x \:=\:\frac{h+5}{\tan44.7^o}\;\;{\color{blue}[2]}

    Equate [1] and [2]: . \frac{h}{\tan39.1^o} \:=\:\frac{h+5}{\tan44.7^o} \quad\Rightarrow . h\tan44.7^o \:=\:(h+5)\tan39.1^o

    . . h\tan44.7^o \:=\:h\tan39.1^o + 5\tan39.1^o \quad\Rightarrow\quad h\tan44.7^o - h\tan39.1^o \:=\:5\tan39.1^o

    . . h(\tan44.7^o - \tan39.1^o) \:=\:5\tan39.1^o \quad\Rightarrow\quad h \:=\:\frac{5\tan39.1^o}{\tan44.7^o - \tan39.1^o}

    Therefore: . h \;=\;22.9693949 \;\approx\; 23.0 m.

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