Question 204104
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You talk about a rope and then a wire. I'm going to go out on a limb and presume that they are the same object.


What you have is a right triangle where the length of the rope, 10 feet, is the hypotenuse.  Let *[tex \Large r] represent the distance from the base of the tower to where the rope meets the ground.  Then *[tex \Large r + 2] must be the height of the tower.


So we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2 + (r + 2)^2 = 10^2 = 100]


Thank you, Mr. Pythagoras.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2 + r^2 + 4x + 4 = 100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r^2 + 4r - 96 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2 + 2r - 48 = 0]


Just solve the quadratic and exclude the negative root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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