SOLUTION: A 100 m radio mast is supported by six cables in two sets of three cables. They are anchored to the ground at an equal distance from the mast. The top set of three cables is attach

Question 1108526: A 100 m radio mast is supported by six cables in two sets of three cables. They are anchored to the ground at an equal distance from the mast. The top set of three cables is attached at a point 20 m below the top of the mast. Each cable in the lower set of three cables is 60 m long and is attached at a height of 30 m above the ground. If all the cables have to be replaced, nd the total length of cable required. Give your answer correct to two decimal places.

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A 100 m radio mast is supported by six cables in two sets of three cables.
They are anchored to the ground at an equal distance from the mast.
The top set of three cables is attached at a point 20 m below the top of the mast.
Each cable in the lower set of three cables is 60 m long and is attached at a height of 30 m above the ground.
If all the cables have to be replaced, and the total length of cable required. Give your answer correct to two decimal places.
:
One set of guy wires (one upper and one lower) have two right triangles which we can solve.
The lower right triangle
a = dist from the pole to guy tie point
b = 30 m
c = 60 m the lower guy wire (the hypotenuse)
a^2 + 30^2 = 60^2
a^2 = 3600 - 900
a^2 = 2700
a =
a = 51.96 m from the pole to the tie point
:
The other triangle using the upper guy wire as the hypotenuse (h)
Tied to a point on the pole 80 m from the ground (20m from the top)
And also 50 m from the tie point
h^2 = 51.96^2 + 80^2
h^2 = 2700 + 6400
h^2 = 9100
h =
h = 95.39 m, the length of the upper guy wire
For one set of guy wires (1 upper & 1 lower): 95.39 + 60 = 155.39 m
For 3 sets of guy wires: 3(155.39) = 466.18 m of wire required
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