I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, $\ce{PH3}$ has nearly pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a nearly pure $s$ orbital.
Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.
While answering this question, I realised that it's not that easy predicting the numbers.
Take the same image of $\ce{B2H6}$:
At first, seeing the 97°, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120° came from, because that is for perfect $sp^2$.
I then realised that I was being stupid and it wasn't that simple--just because you have a 120° doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.
How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?
Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.