SOLUTION: A guy wire is fastened 6 metres from the base of a flagpole and makes angle 30° with the ground. Determine the length of the guy wire and how far up the flagpole it is fastened.

Algebra ->  Trigonometry-basics -> SOLUTION: A guy wire is fastened 6 metres from the base of a flagpole and makes angle 30° with the ground. Determine the length of the guy wire and how far up the flagpole it is fastened.      Log On


   



Question 1115066: A guy wire is fastened 6 metres from the base of a flagpole and makes angle 30° with the ground. Determine the length of the guy wire and how far up the flagpole it is fastened.
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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sin%2830%29=1%2F2 and cos%2830%29=sqrt%283%29%2F2
If r is the length of the guy wire (make drawing to see everything),
then cos%2830%29=6%2Fr
r%2F6=2%2Fsqrt%283%29
r=12%2Fsqrt%283%29
r=%2812%2Fsqrt%283%29%29%28sqrt%283%29%2Fsqrt%283%29%29
r=12%2Asqrt%283%29%2F3
highlight%28r=4%2Asqrt%283%29%29-----length of guy wire.
and Pythagorean theorem could be used for finding the height of pole...



Also recognize, the triangle formed is a special 30-60-90 right triangle. Long leg is 6 meters. Let x be the short leg (height of the flag pole). The wire is then 2x.

x%5E2%2B6%5E2=%282x%29%5E2
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x%5E2%2B36=4x%5E2
3x%5E2=36
x%5E2=12
x=sqrt%282%2A2%2A3%29
highlight%28x=2%2Asqrt%283%29%29------height
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highlight%284%2Asqrt%283%29%29-------guy wire

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A guy wire is fastened 6 metres from the base of a flagpole and makes angle 30° with the ground. Determine the length of the guy wire and how far up the flagpole it is fastened.
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Sketch the picture.
You have a right triangle with base = 6 meters and base angle = 30 degrees.
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cos(30) = base/hypotenuse
Ans: wire = hypotenuse = 6/cos(30) = 6/(sqrt(3)/2) = 12/sqrt(3) = 4*sqrt(3) ft
Ans:: sin(30) = height/hypotenuse
1/2 = height/4sqrt(3)
height = 2sqrt(3) ft
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Cheers,
Stan H.
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