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  1. #1
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    Trig help

    The angle of elevation to a building is 30 degrees. From a point 20 m directly toward the building, the angle of elevation changes to 40 degrees. Find the height of the building. Include a diagram in your solution.

    Do I make 3 triangles for this question?
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  2. #2
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    Quote Originally Posted by sinjid9 View Post
    The angle of elevation to a building is 30 degrees. From a point 20 m directly toward the building, the angle of elevation changes to 40 degrees. Find the height of the building. Include a diagram in your solution.

    Do I make 3 triangles for this question?
    Draw a horizontal line representing the ground. On the left draw a vertical line and on the right put a point 5 inches away from the vertical line.

    label the distance 20m.

    Now you have an angle and distance.

    cos(\theta)=\frac{adj}{hyp}\rightarrow cos(30)=\frac{20}{r}


    cos(\theta)=\frac{adj}{hyp}\rightarrow cos(40)=\frac{20}{r}

    The length you want is y r^2=x^2+y^2
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  3. #3
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    Hello, sinjid9!

    The angle of elevation to a building is 30.
    From a point 20 m closer to the building, the angle of elevation is 40.
    Find the height of the building. Include a diagram in your solution.
    Code:
        A o
          |  *  *
          |     *     *
        h |        *        *
          |           *           *
          |          40 *        30   *
        B o - - - - - - - - o - - - - - - - - o
                   x        D       20        C

    The building is: h \,=\,AB.
    \angle ACB = 30^o,\;\angle ADB \,=\,40^o,\;DC \,=\,20
    Let x \,=\,BD.

    In right triangle ABC\!:\;\;\tan30 \:=\:\frac{h}{x+20} \quad\Rightarrow\quad x \:=\:\frac{h-20\tan30}{\tan30} .[1]

    In right triangle ABD\!:\;\;\tan40 \:=\:\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan40} .[2]


    Equate [1] and [2]: . \frac{h-20\tan30}{\tan30} \;=\;\frac{h}{\tan40} \quad\Rightarrow\quad h\tan40 - 20\tan30\tan40 \;=\;h\tan30

    . . . . . . . . h\tan40 - h\tan30 \:=\:20\tan30\tan40 \quad\Rightarrow\quad h(\tan40-\tan30) \:=\:20\tan30\tan40


    Therefore: . h \;=\;\frac{20\tan30\tan40}{\tan40-\tan30} \;=\;37.01666314 \;\approx\;37\text{ m}

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  4. #4
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    I was thinking of rotating the line from the ground not from the wall.
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