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Math Help - precalc help (law of sines and cosines)

  1. #1
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    precalc help (law of sines and cosines)

    hard to explain without a diagram, but:
    a point on the ground is 200 feet from a water tower
    the angle of elevation to the top of the tower is 18 degrees
    the angle of elevation to the bottom of the tower is 15 degrees
    how tall is the water tower?
    (i think the water tower is on a hill?)
    thanks!
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  2. #2
    Member u2_wa's Avatar
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    Quote Originally Posted by jul07010 View Post
    hard to explain without a diagram, but:
    a point on the ground is 200 feet from a water tower
    the angle of elevation to the top of the tower is 18 degrees
    the angle of elevation to the bottom of the tower is 15 degrees
    how tall is the water tower?
    (i think the water tower is on a hill?)
    thanks!
    Hello jul07010:

    I also think water tower is on the hill

    base=200m

    tan(18)=\frac{Height \color{red}of\color{black} hill+ tower}{200}<br />
    \frac{tan(18)}{200}=Height of hill+ tower

    Height of hill= \frac{tan(15)}{200}

    Height of tower= (Height of hill+ tower)-(Height of hill)
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  3. #3
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    Hello, jul07010!

    I agree with u2_wa: we don't need the Law of Sines . . .


    A point on the ground is 200 feet from a water tower.
    The angle of elevation to the top of the tower is 18.
    The angle of elevation to the bottom of the tower is 15.
    How tall is the water tower?
    (i think the water tower is on a hill?) yes!
    Code:
        B *
          | *
          |   *
        h |     *
          |       *
          |         *
        C *           *
          |   *         *
          |       *       *
        y |     200   *  3 *
          |               *   *
          |               15 * *
        D * - - - - - - - - - - - * A
                      x

    The height of the tower is: h = BC
    Let y = CD,\;x = DA.

    AC is the ground: AC = 200.
    \angle CAD = 15^o,\;\angle BAD = 18^o


    In right triangle CDA:\;\begin{array}{ccccccc}\sin15^o \:=\:\frac{y}{200} &\Rightarrow& y \:=\:200\sin15^o & {\color{blue}[1]}\\<br />
\cos15^o \:=\:\frac{x}{200} & \Rightarrow & x \:=\:200\cos15^o & {\color{blue}[2]} \end{array}

    In right triangle BDA\!:\;\tan18^o \:=\:\frac{h+y}{x} \quad\Rightarrow\quad h \;=\;x\tan18^o - y .[3]

    Substitute [1] and [2] into [3]: . h \;=\;(200\cos15^o)\tan18^o - 200\sin15^o


    Therefore: . h \;\approx\;11 ft.

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