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Math Help - Applications of Right Triangles

  1. #1
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    Applications of Right Triangles

    24.) The angle of elevation of a flagpole from a point level with its base is 45 degrees. from a point 20 ft. further away, the angle of elevation is 30 degrees. What is the hight of the flagpole to the nearest 0.1 ft.?


    I did:

    x*tan(45) = (x + 20)*tan(30)
    x*tan(45) = x*tan(30) + 20tan(30)
    x*tan(45) - x*tan(30) = + 20tan(30)
    x(tan45 - tan30) = 20tan(45)
    x = 20tan(30)/tan(45)-tan
    x = 27.29

    or is it:

    x*tan(30) = (x - 20)*tan(45)
    x*tan(30) = x*tan(45) -20tan(45)
    x*tan(30) - tan(45) = -20tan(45)
    x(tan30 - tan45) = -20tan(45)
    x = -20tan(45)/tan(30) - tan(45)
    x = 47.28

    I don't know which one is right....
    Last edited by lax600; June 2nd 2008 at 11:40 AM. Reason: typed 9 instead of 0
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  2. #2
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    Hello, lax600!

    The problem is easier than you think ... Note that 45 angle!


    The angle of elevation of a flagpole from a point level with its base is 45.
    From a point 20 ft. further away, the angle of elevation is 30.
    What is the height of the flagpole to the nearest 0.1 ft.?
    Code:
        A *
          | * *
          |   *   *
          |     *     *
        x |       *       *
          |         *         *
          |           *           *
          |         45 *         30 *
          * - - - - - - - * - - - - - - - *
          B       x       C      20       D

    Since \angle ACB = 45^o,\;AB = BC = x

    In right triangle ABD\!:\;\;\tan30^o \:=\:\frac{x}{x+20}

    Solve for x\!:\;\;x \;=\;\frac{20\tan30^o}{1 - \tan30^o}

    Since \tan30^o \,=\,\frac{1}{\sqrt{3}}, we have: . x \;=\;\frac{20\cdot\frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \;=\;10(\sqrt{3}+1) \;\approx\;27.3 ft.

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  3. #3
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    so my first answer works... but why not my 2nd answer?
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