Astandard stop sign measures 30 in from flat to flat

And your question is? Or is this a t/f? I'm going to assume you need to find the side length of an octagon?

If you divide this octagon into 8 equal isosceles triangles, then you'll see that each vertex angle is 45 degrees 

If you divide each of those triangles into 2 equal right triangles, then you'll see that each triangle has angles of 22.5 - 67.5 - 90, and one leg length of 15 inches. Now, just use the law of sines to find the length of the other leg. This other leg will be 1/2 of the length of a side 

sin(67.5) / 15 = sin(22.5) / x 
x = 15 * sin(22.5) / sin(67.5) 
x = 15 * sin(45/2) / sin(135/2) 
x = 15 * sqrt((1/2) * (1 - cos(45))) / sqrt((1/2) * (1 - cos(135))) 
x = 15 * sqrt((1/2) * (1 - cos(45)) / ((1/2) * (1 - cos(135))) 
x = 15 * sqrt((1 - cos(45)) / (1 - cos(135))) 
x = 15 * sqrt((1 - sqrt(2)/2) / (1 - (-sqrt(2)/2))) 
x = 15 * sqrt(((2 - sqrt(2)) / 2) / (((2 + sqrt(2))) / 2) 
x = 15 * sqrt((2 - sqrt(2)) / (2 + sqrt(2))) 
x = 15 * sqrt((2 - sqrt(2))^2 / (4 - 2)) 
x = 15 * sqrt(sqrt(2)^2 * (sqrt(2) - 1)^2 / 2) 
x = 15 * sqrt(2 * (sqrt(2) - 1)^2 / 2) 
x = 15 * sqrt((sqrt(2) - 1)^2) 
x = 15 * (sqrt(2) - 1) 

x is one-half of the side 

2x = 30 * (sqrt(2) - 1) 

Now we leave the realm of the exact and head into an approximate realm 

30 * (1.414 - 1) 
30 * 0.414 
3 * 4.14 
12.42 

Is this a statement or question?

Its all  useless we should stop doing everything digitally