= Solution
First obtain the particle speed from kinetic <energy> rather than treating $1\,\mathrm{GeV}$ as its total <energy>. With <Electron> rest <energy> $m_ec^2=0.510999\,\mathrm{MeV}$,
$$
\gamma=1+\frac{1000\,\mathrm{MeV}}{0.510999\,\mathrm{MeV}}\simeq1957.95,\qquad
\beta=\sqrt{1-\gamma^{-2}}\simeq0.999999870.
$$
For <Cherenkov radiation>, the photon propagation direction makes angle $\theta_C$ to the particle trajectory, with $\cos\theta_C=1/(n\beta)$. The speed correction is negligible at three significant digits, but the <Cherenkov photon-cone and wavefront-cone angles> must be distinguished.
If the given opening $\alpha$ is the full cone of photon directions, $\theta_C=\alpha/2=48.75^\circ$. The result is
$$
\boxed{n=\frac1{\beta\cos(\alpha/2)}=1.52\quad\text{(full photon-direction opening)}.}
$$
If instead $\alpha$ denotes the full opening of the conical wavefront, its half-angle is complementary to the photon direction: $\theta_C=90^\circ-\alpha/2=41.25^\circ$. That convention gives
$$
\boxed{n=\frac1{\beta\sin(\alpha/2)}=1.33\quad\text{(full wavefront opening)}.}
$$
The original PDF supplies no cone diagram or definition distinguishing these geometries. Thus the literal full photon-cone interpretation gives $1.52$, while $1.33$ follows from the wavefront interpretation. It would be incorrect to interchange sine and cosine without stating which cone is meant.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2018-q1-cherenkov-conventions.png]
{title=Photon-direction and wavefront cone openings are complementary geometries, giving different interpretations of the supplied angle}
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