More large gaps in Euclid's orchard

The fifth largest pair of adjacent gaps in this range are on either side of what angle up from the x-axis?

If use the same Python script shown earlier in the Classic answer, using the value of $k=6$ in the Python code, we get that for large enough values of $R,$ the next few largest pair of adjacent gaps occur at the following bearings, respectively: \begin{align*}\theta_2^* &= \tan^{-1} \left(\frac{1}{3} \right) \approx 18.434948823\dots^\circ \\ \theta^*_3 &= \tan^{-1} \left(\frac{2}{3}\right) \approx 33.690067526\dots^\circ \\ \theta^*_4 &= \tan^{-1} \left(\frac{1}{4}\right) \approx 14.036243468\dots^\circ.\end{align*} Through now, we still maintain that these are corresponding to the next highest peaks of Thomae's function at $x^*_2 = \frac{3}{4}, x^*_3 = \frac{3}{5}$ and $x^*_4 = \frac{4}{5},$ respectively. However, continuing one step further, we see that for sufficiently large values of $R$, the fifth largest pair of adjacent gaps occurs at $$\theta^*_5 = \tan^{-1} \left(\frac{3}{4}\right) \approx 36.869897649\dots^\circ,$$ which in fact skips over the next highest Thomae's function value of $\tilde{x} = \frac{5}{6}$ in favor of $x^*_5 = \frac{4}{7}.$ Don't feel too bad for $\tilde{x} = \frac{5}{6},$ since it is equivalent to the bearing $\theta^*_6 = \tan^{-1} \left(\frac{1}{5}\right) \approx 11.309932474\dots^\circ,$ which is in fact the bearing around which you can find the next largest adjacent pair gaps. Perhaps you could still feel bad for Thomae's function's explanatory powers, though.