Answer: E

Small Hint:

Let $\alpha$ and $\beta$ be the half-central angles subtended by sides $3$ and $5$

Big Hint:

Use $\alpha+\beta=60^\circ$ and the chord ratio to find $\sin^2\alpha$, then apply $\frac{\sin3\alpha}{\sin\alpha}$

Solution:

Let the circle have radius $R,$ and let $\alpha,\beta$ be the half-central angles for the sides $3,5.$ Then $3=2R\sin\alpha,$ $5=2R\sin\beta,$ and $\alpha+\beta=60^\circ.$ Thus $\frac53=\frac{\sin(60^\circ-\alpha)}{\sin\alpha}$ $=\frac{\sqrt3}{2}\cot\alpha-\frac12,$ so $\tan\alpha=\frac{3\sqrt3}{13}$ and $\sin^2\alpha=\frac{27}{196}.$ The dividing chord spans the three consecutive sides of length $3,$ so $\frac L3=\frac{\sin3\alpha}{\sin\alpha}$ $=3-4\sin^2\alpha=3-\frac{27}{49}=\frac{120}{49}.$ Therefore $L=\frac{360}{49},$ so $m+n=409.$ The correct answer is E.