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Question 1197645: A 200 ft cell phone tower will be built somewhere on the west side of an 800 ft tall hill that is 2800 feet wide from west to east at the base. How far up the hill must the tower be placed to provide a signal to anyone on the east side of a 600 ft lake? (Note: the lake starts at the base of the hill.)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
For the sake of simplicity, I'll assume the hill is an isosceles triangle such that the two congruent sides are slanted. The triangle points directly upward as shown in the diagram below. Triangle AEG is the hill.
A = base of the hill on the west side
B = base of the cell tower
C = top of the cell tower
D = point directly below B, same level as point A
E = point at the top of the hill
F = point directly below E, same level as point A
G = base of the hill on the east side
H = point 600 ft east of G
Segment GH represents the lake.
Triangle AEG is the hill.
AE = EG are the two congruent slanted sides of isosceles triangle AEG.
Points C, E, and H are collinear.
Some useful lengths or distances to keep in mind
AG = 2800
EF = 800
GH = 600
The goal is to have a direct line of sight from point C to point H.
If point B is too low on the hill, then segment CH will cut into the mountain, which means a cell signal will get blocked.
Let
x = length of segment AD
y = length of segment BD
Since triangle AEG is isosceles, and point F is the midpoint of AG, this means AF = AG/2 = 2800/2 = 1400.
It also means FG = 1400 as well.
Turn your attention to right triangles AFE and ADB. They are similar triangles. We can use the AA (angle angle) similarity theorem to prove this claim.
The similar triangles allow us to form the proportion shown below and we'll solve for x.
AD/BD = AF/EF
x/y = 1400/800
x/y = 14/8
x/y = 7/4
4x = 7y
x = 7y/4
x = 1.75y
This means AD = x = 1.75y
and
DF = AF-AD
DF = 1400-1.75y
Both DF and BD are algebraic expressions in terms of y.
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Now turn your attention to right triangles CDH and EFH.
Using the AA similarity theorem, we can prove them to be similar triangles.
From there we say
CD/DH = EF/FH
(CB+BD)/(DF+FG+GH) = EF/(FG+GH)
(200+y)/(1400-1.75y+1400+600) = 800/(1400+600)
(200+y)/(3400-1.75y) = 800/2000
(200+y)/(3400-1.75y) = 2/5
Let's cross multiply and solve for y.
(200+y)/(3400-1.75y) = 2/5
5(200+y) = 2(3400-1.75y)
1000+5y = 6800-3.5y
5y+3.5y = 6800-1000
8.5y = 5800
y = 5800/8.5
y = 682.35294117647
which is approximate.
Now we can find x.
x = 1.75y
x = 1.75*682.35294117647
x = 1194.11764705882
which is also approximate
Lastly, use the pythagorean theorem to find the length of hypotenuse AB of right triangle ADB.
a = 1194.11764705882
b = 682.35294117647
c = sqrt(a^2+b^2)
c = sqrt(1194.11764705882^2+682.35294117647^2)
c = 1375.32632176858
Like the others, this is approximate.
Let's say we rounded that to four decimal places.
That gives roughly 1375.3263
If you started at point A, and moved about 1375.3263 ft up the hill, then you'll arrive at a location where the top of the tower (point C) can establish a line of sight with point H. Anything higher than this point will provide a better line of sight.
So it might be better to round it to say the next largest integer to get 1376 ft up the hill. We'll slightly overshoot the goal but it's better to overestimate here than come up short.
We could go to the very top of the hill and avoid all of this math altogether. But 1376 ft represents the lowest amount of integer feet climbing the hill, such that we can get a line of sight. Meaning that if it costs more money the higher you climb, then it's in the company's best interests to set up the tower at point B rather than at point E.
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Answer: Approximately 1376 ft up the hill
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