Comparing bond angles in carbonyl dichloride and carbonyl dibromide Background
I was studying VSEPR and hybridization theory and one of the popular questions in many textbooks was comparison of bond angle of $\ce{NH3}$, $\ce{NF3}$ and $\ce{NCl3}$.
$\ce{NF3}$ had smaller bond angle as electrons were nearer to the more electronegative fluorine atom, thereby reducing bond pair - bond pair repulsion (of N-F bonds) and thus allowing contraction of bond angle.
Further, in $\ce{NCl3}$ bond angle was larger than both $\ce{NH3}$ and $\ce{NF3}$ due to steric repulsion between large Cl atoms. So the order was $\ce{NCl3}$&gt;$\ce{NH3}$&gt;$\ce{NF3}$.
Now, a question that I later saw in one of my textbooks was the analysis of bond angles of $\ce{COCl2}$ and $\ce{$\ce{COBr2}$}$, which have trigonal planar geometry. According to my textbook, $\ce{COCl2}$ has smaller $\ce{O-C-Cl}$ angle, less than $\mathrm{120^{\circ}}$, (and thereby greater $\ce{Cl-C-Cl}$ angle). The angles for $\ce{COBr2}$ were not given.
My Approach
Now if we apply the reasoning given for $\ce{NCl3}$ and $\ce{NH3}$, $\ce{NF3}$ in this question comparing $\ce{COCl2}$ and $\ce{COBr2}$, we would find that the $\ce{C-X}$ bond pair electrons would be more polarized towards X, if X is Cl, and thus analogously to $\ce{NF3}$, it should contract leading to a smaller bond angle than in $\ce{COBr2}$.
Also, due to steric repulsion, $\ce{C-X}$ bond angle in $\ce{COBr2}$ should be greater than $\ce{COCl2}$.
Solution given in the book
The author has stated that there are four electrons in the $\ce{C-O}$ bond
Thus, bond pair repulsion between electrons of $\ce{C-O}$ bond and $\ce{C-X}$ bond would be much greater than any repulsion between the electrons of the $\ce{C-X}$ bonds.
As electrons in the $\ce{C-X}$ bond would be closer to carbon atom (and thus $\ce{C-O}$ bond pair electrons) in $\ce{COBr2}$, therefore a greater amount of repulsion would be present between them. This repulsion would lead to contraction of Br-C-Br bond angle leading to an angle less than that of $\ce{COCl2}$.
However, no data was provided in support or against it. Also no reasoning was given why they would be greater or less than $\mathrm{120^{\circ}}$, only the above reasoning for comparing the bond angle was provided.
I have not been able to find any bond angle data. My teachers are also not sure. Also, the textbook I referred to is infamous for having lots of errors.
Which reasoning is correct, or is it something else entirely?
What is the actual data?
The reasoning you utilized for comparing the bond angles of $\ce{NH3}$, $\ce{NF3}$, and $\ce{NCl3}$, while sensible, is simplistic and works best only for a few cases where weighing the factors of steric repulsion against bond pair-bond pair repulsions is feasible. It seems that you have encountered this very problem in the case of comparing the bond angles of $\ce{COCl2}$ and $\ce{COBr2}$.
To circumvent these issues, a better method(less prone to failures) would be using Bent's rule. Now, there has been a lot of discussion on this website over what this rule actually is (see here) and why it is physically useful (see this). The Wikipedia article on the same has a wonderful intuitive discussion under the &quot;Justification&quot; tab for why this rule exists. However, I will not be delving into the origins of this rule, and just state outright that the basic premise of this rule is that not all hybrid orbitals are equivalent under a given hybridization scheme. The percentage s-character and p-character (that is, the amount of s-orbital or p-orbital present in a given hybrid orbital as a percentage of the whole) varies as per the electronegativities of the substituents.
A statement of this rule was provided by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

The idea is that the amount of p character will increase (alternatively, the amount of s character will decrease) in those bonds which have more electronegative substituents (that is, the atom other than carbon in the bond), while simultaneously the amount of p character will decrease(alternatively, the amount of s character will increase) in bonds towards less electronegative substituents in such a manner that the net sum of percentage s and p characters over all the bonds in the molecule comes out to be $100$%. This is essentially a redistribution of the orbital energies across bonds to minimize the energy of the system. Further details on this rule can be found in the links mentioned above.
Once you are aware of this rule, applying it to your case becomes a fairly simple task. I will try to compare the $\ce{Cl-C-Cl}$ and $\ce{Br-C-Br}$ angles here, the orders for the $\ce{O-C-Cl}$ and $\ce{O-C-Br}$ angles can be easily inferred from there. From a classical hybridization perspective, the carbon in both these molecules uses $\ce{sp^2}$ hybrid orbitals for bonding, and all the three hybrid orbitals thus generated should possess equivalent atomic orbital's character ($33.33$% s character and $66.66$% p character approximately). But as per Bent's rule, since $\ce{Cl}$ and $\ce{Br}$ are both more electronegative than carbon, they will increase p character towards their direction (that is, in the $\ce{C-Cl}$ and $\ce{C-Br}$ bonds respectively). Since $\ce{Cl}$ is even more electronegative than $\ce{Br}$, the p character increase in the $\ce{C-Cl}$ bonds will be more than that what will happen in the $\ce{C-Br}$ bonds.
Hence, we can say that more p character will be directed in the two $\ce{C-Cl}$ bonds as compared to the two $\ce{C-Br}$ bonds. This directly bears a relation to the bond angle: more is the p character in the constituent bonds, lesser will be the bond angle(you can see this from a fairly easy observation: $\mathrm{sp}$ hybrid orbitals are generally at an angle of $180°$ with $50$% p character, the $\mathrm{sp^2}$ hybrid orbitals are generally at an angle of $120°$ with $66.66$% p character and the $\mathrm{sp^3}$ hybrid orbitals are generally at an angle of $109.5°$ with $75$% p character).
Hence, Bent's rule predicts that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ should be lesser than the $\ce{Br-C-Br}$ of $\ce{COBr2}$. Experimental data says that the $\ce{Cl-C-Cl}$ bond angle of $\ce{COCl2}$ is $111.83 ± 0.11°$, while the $\ce{Br-C-Br}$ of $\ce{COBr2}$ is $112.3 ± 0.4°$. Hence, the prediction made by Bent's rule is confirmed.
Notice that the difference between the given bond angles is quite small (around $1°$). This is also somewhat expected, as the electronegativity difference between chlorine and bromine is also quite small (approximately $0.2$).
The advantage of employing this method is that you don't have to consider conflicting factors like those you stated in your approach. The rule explains these angle variations based on Coulson's theorem, from which the following relation pops out:

$$\cos \theta = \frac{S}{S-1}=\frac{P-1}{P}$$
for $\theta \in (90^\circ,180^\circ)$

($\theta$ stands for the bond angle while ${P}$ and ${S}$ stand for p character and s character respectively)
As you can clearly see, when the p character goes up, the value of $\cos \theta$ also goes up if you closely observe the simplified form $\cos \theta = 1 - \frac{1}{P}$. This means that the value of $\theta$ will decrease for the given range of $\theta$ (as $\cos \theta$ is a decreasing function for $\theta \in (90^\circ,180^\circ)$)
Hence, you can generally predict things quite accurately with Bent's rule without getting confused with contrasting factors, as they are usually taken care of while determining the atomic orbital characters.