Bond character and bond angles
S orbitals are lower in energy than P orbitals.  Electrons prefer to be in as low an energy orbital as possible.  Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride).  The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond.  So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it?  Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them.  It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted).  The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).