Two telephone poles are 60ft. apart, and one of them is 6ft. higher than the other. If the length of the wire between the poles is 61.3 ft., how much wire was allowed for sag?.
the straight-line distance is
√(60^2+6^2) = 60.3 ft, so ...
To determine the amount of wire allowed for sag, we need to find out the difference in height between the two poles.
Let's assume that the lower pole's height is 'x' ft. This means the higher pole's height would be 'x + 6' ft.
We can create a right-angled triangle to represent the situation. The distance between the poles will be the base of the triangle, which is 60 ft. The difference in height between the poles will be the height of the triangle.
Using Pythagoras' theorem, we can find the height as follows:
(height)^2 + (base)^2 = (hypothesis)^2
(x)^2 + 60^2 = 61.3^2
Simplifying the equation:
x^2 + 3600 = 3751.69
x^2 = 3751.69 - 3600
x^2 = 151.69
Taking the square root of both sides:
x = √151.69
x ≈ 12.31 ft
Therefore, the difference in height between the poles is approximately 12.31 ft. Since the wire length is 61.3 ft, the amount of wire allowed for sag is 61.3 - 12.31 = 48.99 ft.