The height of the flagpole to the nearest inch is 388 inches.
Explanation
According to the below diagram, AB is the flagpole which is secured on opposite sides by two guy wires, AB and AC.
Suppose, the length of the flagpole(AB) is  [tex]x[/tex] ft.
As each of the guy wires is 5 ft longer than the pole, that means [tex]AC= AD= (x+5)ft[/tex]
Also given that, the distance between the points where the wires are fixed to the ground is equal to the length of one guy wire. That means, [tex]CD= (x+5)ft[/tex]
As [tex]\triangle ABC[/tex] and [tex]\triangle ABD[/tex] are congruent to each other, so [tex]BC=BD= \frac{1}{2}CD= \frac{1}{2}(x+5) ft[/tex]
Now in right triangle [tex]ABD[/tex], using Pythagorean theorem......
[tex]AB^2+BD^2= AD^2\\ \\ x^2+[\frac{1}{2}(x+5)]^2= (x+5)^2\\ \\ x^2+ \frac{1}{4}(x+5)^2 = (x+5)^2\\ \\ 4x^2+(x+5)^2= 4(x+5)^2\\ \\ 4x^2= 3(x+5)^2\\ \\ 4x^2=3(x^2+10x+25)\\ \\ 4x^2-3x^2-30x-75=0\\ \\ x^2-30x-75=0[/tex]
Using quadratic formula, we will get.......
[tex]x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1)(-75)}}{2(1)}\\ \\ x= \frac{30\pm \sqrt{1200}}{2}\\ \\ x= 32.3205..., x=-2.3205...[/tex]
(Negative value is ignored as the height of the pole can't be in negative)
So, the height of the flagpole will be: Â [tex]32.3205... ft =(32.3205...\times 12)inches = 387.846... \approx 388 inches[/tex]
