A 24 meters pole is held by a three guy wire into vertical position. Two of the guy wires are of equal length. The third wire is 5 meters longer than the other two and is attached to the ground 11 meters farther from the foot of the pole than the two equal wires. Find the length of the wire.

Answer 1

Let the shorter wire be x m long and let that wire be y m from the base of the pole
so x^2= 576 + y^2 ----> x^2 - y^2 = 576

let the longer be x+5 and we have
(x+5)^2 = 24^2 + (y+11)^2
x^2 + 10x + 25 = 576 + y^2 + 22y + 121
x^2 - y^2 = 22y - 10x + 672

22y - 10x + 672 = 576
22y - 10x = - 96
y = (10x-96)/10

sub this into
x^2 - y^2 = 576
x^2 - (100x^2 - 1920x + 9216)/100 = 576
times 100
100x^2 - 100x^2 + 1920x - 9216 = 57600
1920x = 66816
x = 34.8 , then y = 25.2

check:
x^2 - y^2 = 576
LS = 34.8^2 - 25.2^2 = 576 = RS

also checks out for the 2nd equation, I will let you verify that.

Answer 2

A wire extends from the top of a 12-foot light pole to the ground. The wire forms an angle of 53°with the ground. Find the length of the wire to the nearest tenth.