solution
For this case we can do the following:
[tex]AB=2\pi r(\frac{x}{360})=2\pi\cdot9in(\frac{40}{360})=2\pi=6.283\text{ in}[/tex]
And then the lenght of the ARC ACB would be:
[tex]\text{ACB}=2\pi(9)-2\pi=2\pi(9-1)=16\pi=50.265in[/tex]
Final answer 16 pi
