Okay, here we have this:
Considering the provided information, and the following chord length equation, we obtain the following:
[tex]\begin{gathered} K=2r\cdot sen(\frac{\theta}{2}) \\ 24.5=2r\cdot\text{sen(}\frac{57.5}{2}\text{)} \end{gathered}[/tex]
Now, let's solve for r:
[tex]\begin{gathered} 2r\sin \mleft(\frac{57.5^{\circ\:}}{2}\mright)=24.5 \\ r=\frac{24.5}{2\sin\left(28.75^{\circ\:}\right)} \\ r=25.5\text{ in} \end{gathered}[/tex]
Finally we obtain that the radius of the circle is approximately 25.5 inches.
