= Solution
The observer is static at $r_0$, with <four-velocity> $U=f_0^{-1/2}\partial_t$, where $f_0=1-2m/r_0$. Angles are measured in the observer's <rest frame>, using the positive spatial <metric tensor>
$$
d\ell_{\mathrm{space}}^2=f_0^{-1}dr^2+r_0^2d\phi^2
$$
on the equatorial plane. One cannot directly normalize a photon tangent with the full <Lorentzian metric> to define an angle, because that tangent is null.
At the observer the spatial direction of an outward-moving grazing ray is proportional to $w=A\partial_r+\partial_\phi$, with $A=(dr/d\phi)_{r_0}$. Its angle $\psi=\alpha/2$ to the outward radial direction equals the corresponding sky angle to the inward center of the image. Taking the spatial <inner product> with $e_r=\partial_r$,
$$
\cos\psi
=\frac{|\langle w,e_r\rangle|}
{\sqrt{\langle w,w\rangle\langle e_r,e_r\rangle}}
=\frac{|A|}{\sqrt{A^2+f_0r_0^2}}.
$$
Consequently
$$
\boxed{
\cos\frac\alpha2
=\left(1+\frac{f_0r_0^2}{A^2}\right)^{-1/2}.
}
$$
The limiting value $A=0$ is understood by continuity.
For a grazing ray the <areal radius> has a turning point at the stellar surface $R$, so $\dot r=0$ there. The radial equation gives
$$
C=\frac{f(R)}{R^2}.
$$
Since $R=4m>3m$, the surface lies outside the <photon sphere>. The function $f/r^2$ decreases for $r>3m$, so this tangential ray can escape to every $r_0>R$ and correctly defines the image's rim. For the specified surface,
$$
f(R)=\frac12,\qquad
C=\frac1{32m^2},\qquad
b=\frac R{\sqrt{f(R)}}=4\sqrt2\,m.
$$
Also $dr/d\phi=-r^2\,du/d\phi$, hence
$$
\boxed{
A^2=r_0^4\left(\frac1{32m^2}-\frac{f_0}{r_0^2}\right),
\qquad
A=\pm r_0\sqrt{\frac{r_0^2}{32m^2}-f_0}.
}
$$
Its sign depends on the orientation of the ray and of $\phi$, and does not affect the <angular diameter>. Substitution gives
$$
\boxed{
\cos\frac\alpha2
=\sqrt{1-\frac{32m^2}{r_0^2}\left(1-\frac{2m}{r_0}\right)},
\qquad
\alpha=2\arcsin\left[
\frac{4\sqrt2\,m}{r_0}\sqrt{1-\frac{2m}{r_0}}
\right].
}
$$
This <apparent angular radius of a spherical star in Schwarzschild spacetime> has useful limiting checks. As $r_0\downarrow R$, the star occupies a hemisphere, $\alpha\to\pi$. Far away, $\alpha\simeq2b/r_0=8\sqrt2\,m/r_0$: <Schwarzschild light deflection> increases the apparent radius by a factor $\sqrt2$ relative to the flat-space value $R/r_0$.
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