= Solution
For $l=2$, $Q=2.7925\,\mathrm{\mathring A^{-1}}$. The scattering triangle gives
$$
\cos\phi=\frac{k_i^2+k_f^2-Q^2}{2k_ik_f},\qquad\boxed{|\phi|=46.51^\circ.}
$$
To state signed sample angles without an implicit axis convention, take the incident beam along $+x$, positive angular <rotation> counterclockwise, and at $\omega=0$ choose $+a$ along $+x$ and $+c$ along $+y$. Then
$$
\mathbf k_i=k_i(1,0),\quad\mathbf k_f=k_f(\cos\phi,\sin\phi),\quad
\hat{\mathbf c}=(-\sin\omega,\cos\omega).
$$
The positive-$l$ condition is the vector equation $\mathbf k_i-\mathbf k_f=Q\hat{\mathbf c}$. Put
$$
\beta=\tan^{-1}\frac{k_f\sin|\phi|}{k_i-k_f\cos\phi}=34.80^\circ.
$$
The two detector sides then give, modulo full <rotations>,
$$
\boxed{(\omega,\phi)=(-55.20^\circ,-46.51^\circ)
\quad\text{or}\quad(-124.80^\circ,+46.51^\circ).}
$$
These pairs obey the stated same-sense sign convention and transfer <momentum> along $+c$. The lower-right panel below illustrates the first choice. If the initial $+c$ direction is chosen on the opposite side of $+a$, a commonly used positive-angle pair is $(55.20^\circ,46.51^\circ)$. Merely taking absolute values without fixing that <crystal>-axis convention can exchange $l=+2$ and $l=-2$. The measurable scattering angle and the vector-triangle construction are invariant.
\Image[/past-exam-of-the-condensed-matter-physics-major-option-course-of-the-university-of-oxford/2018-lattice-and-band-geometry.png]
{title=Honeycomb geometry, reciprocal-space hexagon, sublattice band gap and a signed neutron-scattering triangle}
{description=Original calculations and schematic geometry. Lattice panels use $a=1$; the band panel compares normalized two-sublattice models. The neutron panel uses the explicit signed-axis convention of Question 1(c)(ii).}
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