Cyclic quadrilateral $ABCD$ has lengths $BC=CD=5$ and $DA=8$ with $\angle CDA=60^\circ$. What is the length of the longer diagonal of $ABCD$?

1. In $\triangle CDA$, $AC^2=5^2+8^2-2\cdot5\cdot8\cos60^\circ=25+64-40=49$, so $AC=7$.
2. $ABCD$ is cyclic, so $\angle ABC=180^\circ-60^\circ=120^\circ$. In $\triangle ABC$: $7^2=AB^2+5^2-2\cdot AB\cdot5\cos120^\circ$, i.e. $AB^2+5AB-24=0$, so $AB=3$.
3. By Ptolemy's theorem, $AC\cdot BD=AB\cdot CD+BC\cdot DA=3\cdot5+5\cdot8=55$, so $BD=\dfrac{55}7\approx7.9>7$.
4. The longer diagonal is $BD=\boxed{\dfrac{55}7}$.