= Solution
The subbranch spacing gives $\widetilde A-\widetilde B=4.9\ \mathrm{cm^{-1}}$, hence \b[$\widetilde A\simeq5.15\ \mathrm{cm^{-1}}$]. Rotation about the C–I <symmetry> axis has moment
$$
I_a=\frac{h}{8\pi^2c\widetilde A}\simeq5.44\times10^{-47}\ \mathrm{kg\,m^2}.
$$
C and I lie on this axis, so only the three <hydrogens> contribute to $I_a$. If $\theta$ is the I–C–H angle, their perpendicular distances are $r_{\rm CH}\sin\theta$. Taking integer <hydrogen> mass gives
$$
I_a=3m_Hr_{\rm CH}^2\sin^2\theta,\qquad
\sin^2\theta\simeq\frac{5.44\times10^{-47}}{3(1.66054\times10^{-27})(1.1\times10^{-10})^2}\simeq0.902.
$$
The molecular arrangement selects the obtuse solution, so \b[$\theta\simeq108^\circ$]. The supplementary acute angle describes the opposite axial direction, not the bond angle to iodine. This is a rigid-geometry estimate based on an effective <rotational constant> and the stipulated bond length, rather than a precision equilibrium-structure measurement.
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