



 
Hi Lauchie, In my diagram below the octagon is divided into 8 triangles, each with a vertex at the center $C$ of the octagon. The 8 angles at $C$ have the same measure and hence the measure of each is $\large \frac{360^o}{8} \normalsize = 45^{0}.$ Using the facts that the triangle $ABC$ is isosceles and that the sum of the angle measures in a triangle is $180^{o}$ you can show that the angle $CAB$ measures $67.5^{o}.$ Hence each of the angle cuts is $67.5^{o}.$ I redrew the diagram and added a point $D$ midway between $A$ and $B.$ You said that the pool is a 24 foot round pool so the diameter of the pool is 24 feet and its radius in thus 12 feet. When you put the octagonal deck around the pool I don't know if you want $AC,$ the length of the line from $A$ to $C$ to be 12 feet or $DC.$ the length of the line from $D$ to $C$ to be 12 feet. DC = 12 feet: \[\tan\left(CAD\right) = \tan\left(67.5^o\right) = 2.414 = \frac{DC}{AD} \] so \[AD = \frac{12}{2.414} = 4.97 \mbox{ feet.}\] and thus $AB = 2 \times 4.97 = 9.9$ feet which is 9 feet 9 inches. AC = 12 feet: \[\cos\left(CAD\right) = \cos\left(67.5^o\right) = 0.383 = \frac{AD}{AC} \] so \[AD = 0.383 \times 12 = 4.59 \mbox{ feet.}\] and thus $AB = 2 \times 4.59 = 9.2$ feet which is 9 feet 2 inches. I hope this works for you, 



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