= Solution
Because <optical dispersion> makes the <refractive index> depend on <wavelength>, no single <angle of incidence> makes every component of white <light> travel parallel to the base. Choose the supplied central visible <wavelength> $535\,\mathrm{nm}$, with $n_0=1.86$, as the design component. <Snell's law> gives
$$
\boxed{i_0=\arcsin[1.86\sin(27.5^\circ)]=59.188^\circ,\qquad
\delta_0=2i_0-55^\circ=63.376^\circ.}
$$
Keep this <angle of incidence> fixed for all colours. The other components have
$$
r_1(n)=\arcsin\frac{\sin i_0}{n},\qquad
r_2(n)=A-r_1(n),\qquad
e(n)=\arcsin[n\sin r_2(n)],\qquad
\delta(n)=i_0+e(n)-A.
$$
Substitution gives
$$
\begin{array}{c|c|c|c}
\lambda\ (\mathrm{nm})&n&r_1&e\\\hline
370&1.94&26.277^\circ&68.800^\circ\\
535&1.86&27.500^\circ&59.188^\circ\\
700&1.83&27.990^\circ&56.210^\circ
\end{array}
$$
All three emerge: the largest value of $n\sin r_2$ is about $0.9323<1$. The angular spread between the extreme supplied <wavelengths> is
$$
\boxed{\Delta\theta=e(1.94)-e(1.83)=12.590^\circ\simeq0.21973\,\mathrm{rad}.}
$$
The respective deviations are $72.988^\circ$, $63.376^\circ$ and $60.398^\circ$. Applying the symmetric formula independently to the two extreme <refractive indices> would instead give $72.221^\circ-60.344^\circ=11.877^\circ$; that compares differently oriented <optical prisms>, not the emerging components of one fixed white-light beam. It can be a rough approximation but misses the stated fixed-incidence geometry. The supplied $370\,\mathrm{nm}$ value is near <ultraviolet>, so the computed spread refers to the entire supplied band, not a universally defined visible-white spectrum.
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