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Geometry Problem

Ernesto223

New member
Joined
Dec 17, 2012
Messages
1
The Vriend Monument is a sacred place for Geometry students in Western Albania. Chance Jones uses a surveying device (set on ground) to measure the angle of elevation to the top of the monument to be 3.2 degrees. He then walks one mile (5,280 feet) closer and measures the angle of elevation to be 15.4 degrees. How high (in feet) is the monument?

If someone can could you give me a full explanation? Thank you in advanced. :D
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, Ernesto223!

The Vriend Monument is a sacred place for Geometry students in Western Albania.
Chance Jones measures the angle of elevation to the top of the monument to be 3.2 [SUP]o[/SUP].
He then walks one mile (5,280 feet) closer and the angle of elevation is 15.4 [SUP]o[/SUP].
How high (in feet) is the monument?

Code:
    V *
      | * *
      |   *   *
    h |     *     *
      |        *       *
      |         *         *
      |     15.4[SUP]o[/SUP] *      3.2[SUP]o[/SUP] *
    M *-------------*-------------* A
      : - -  x  - - B - - 5280  - :
The height of the monument is: \(\displaystyle h = VM.\)

Jones stands at \(\displaystyle A\) and \(\displaystyle \angle VAM = 3.2^o.\)
He moves to \(\displaystyle B\!:\;AB = 5280\text{ ft},\;\angle VBM = 15.4^o.\)
Let \(\displaystyle x = MB.\)

In \(\displaystyle \Delta VMA\!:\;\tan3.2^o \:=\:\dfrac{h}{x+5280} \quad\Rightarrow\quad x \:=\:\dfrac{h-5280\tan3.2^o}{\tan3.2^o}\) .[1]

In \(\displaystyle \Delta VMB\!:\;\tan15.4^o \:=\:\dfrac{h}{x} \quad\Rightarrow\quad x \:=\:\dfrac{h}{\tan15.4^o}\) .[2]

Equate [1] and [2]: .\(\displaystyle \dfrac{h-528\tan32^o}{\tan3.2^o} \:=\:\dfrac{h}{\tan15.4^o} \)

. . \(\displaystyle h\tan15.4^o - 5280\tan3.2^o\tan15.4^o \:=\:h\tan3.2^o\)

. . \(\displaystyle h\tan15.4^o - h\tan3.2^o \;=\;5280\tan3.2^o\tan15.4^o\)

. . \(\displaystyle h(\tan15.4^o - \tan3.2^o) \;=\;5280\tan3.2^o\tan15.4^o\)

. . \(\displaystyle h \;=\;\dfrac{5280\tan3.2^o\tan15.4^o}{\tan15.4^o - \tan3.2^o} \;=\;370.3747019\)


Therefore: .\(\displaystyle h\;\approx\;370.4\text{ ft}\)
 

CoCoD

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Joined
Dec 19, 2012
Messages
1
Thanks

Hay thanks i really needed that but how would i draw that picture that would help me a lot.:D
 
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