Trig help

Apr 2010
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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?
 

dwsmith

MHF Hall of Honor
Mar 2010
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Florida
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.

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


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

The length you want is y \(\displaystyle r^2=x^2+y^2\)
 

Soroban

MHF Hall of Honor
May 2006
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6,341
Lexington, MA (USA)
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: \(\displaystyle h \,=\,AB.\)
\(\displaystyle \angle ACB = 30^o,\;\angle ADB \,=\,40^o,\;DC \,=\,20\)
Let \(\displaystyle x \,=\,BD.\)

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

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


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

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


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

 
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dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
I was thinking of rotating the line from the ground not from the wall.