Question

A flagpole is secured on opposite sides by two guy wires, each of which is 5 ft longer than the pole. The distance between the points where the wires are fixed to the ground is equal to the length of one guy wire. How tall is the flagpole (to the nearest inch)?

Answer 1

You have an equalateral triangle of x+5 on each side.
Using the pythagorean formula where (x+5) is the hypotenuse,
one side is x the height of the pole, and (x+5)/2, which is the side opposite the 30 degree angle, or 1/2 the hypotenuse, 
(x=5)^2 = x^2 + ((x+5)/2)^2 
solving for x = 5/(.414) = 12.077' or 144.9" 

hope this will help you :-))

Answer 2

$\huge\boxed{SOLUTION}$

The height of the flagpole to the nearest inch is 388 inches.

Further 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 xx ft.

As each of the guy wires is 5 ft longer than the pole, that means AC= AD= (x+5)ftAC=AD=(x+5)ft

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, CD= (x+5)ftCD=(x+5)ft

As ABC and ABD are congruent to each other, so BC=BD= $\frac{1}{2}CD= \frac{1}{2}$

Now in right triangle ABDABD , using Pythagorean theorem

$\begin{lgathered}AB^2+BD^2= AD^2\\ \\ x^2+[\frac{1}{2}(x+5)]^2= (x+5)^2\\ \\ x^2+ \frac{1}{4}(x+5)^2 = (x+5)^2\\ \\ 4x^2+(x+5)^2= 4(x+5)^2\\ \\ 4x^2= 3(x+5)^2\\ \\ 4x^2=3(x^2+10x+25)\\ \\ 4x^2-3x^2-30x-75=0\\ \\ x^2-30x-75=0\end{lgathered}$

Using quadratic formula, we will get

$\begin{lgathered}x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1)(-75)}}{2(1)}\\ \\ x= \frac{30\pm \sqrt{1200}}{2}\\ \\ x= 32.3205..., x=-2.3205...\end{lgathered}$

(Negative value is ignored as the height of the pole can't be in negative)

So, the height of the flagpole will be: 32.3205... ft =(32.3205...$\times 12)inches = 387.846...$ $\approx 388$ inches32.3205...ft=(32.3205...×12)inches=387.846...≈388inches