Nucleophilic attack at a carbonyl - angle of attack
Let's take the following approach to answer your interesting question:


Use the bent bond model to describe the carbonyl
Calculate the bond angles around the carbonyl carbon
Treat the reaction as an $\mathrm{SN2}$ reaction
Use trigonometry to deduce the angle of nucleophilic approach


The bent-bond model is widely used in chemistry to describe highly-strained systems such as cyclopropane and double bonds (ref_1, ref_2).  In the case of a double bond, we wind up with a two-membered ring (which nicely explains the high heat of formation of such compounds). Here we will use it to describe the carbonyl double bond.  Here is a picture of what the bent bond description of a carbonyl looks like (it's the top left diagram that's of interest, note too that it is a picture of an amide carbonyl).



(image source)

In a simple carbonyl compound like formaldehyde the $\ce{H-C-H}$ bond angle is around 118°.  Using Coulson's theorem (ref_2, ref_4) we can use this bond angle to estimate the hybridization of the carbonyl $\ce{C-H}$ bonds.

$$\ce{1+\lambda^2cos(118)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the $\ce{C-H}$ bond. From this equation we find that $\ce{\lambda^2=}$ $\mathrm{2.13}$, or the carbonyl $\ce{C-H}$ bonds are $\ce{sp^{2.13}}$ hybridized.

We can now sum up the s-character in the 4 bonds around the carbonyl carbon (it must add up to 1) and deduce the hybridization of the 2 $\ce{C-O}$ bent bonds.

$$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-O}^2}=1}$$ 

Solving this equation we find that $\ce{C-O}$ bent bonds are $\ce{sp^{4.54}}$ hybridized.

Substituting this information back into Coulson's theorem allows us to estimate the $\ce{O-C-O}$ bond angle as being equal to 102.7°.

In an $\mathrm{SN2}$ reaction the nucleophile approaches 180° from the leaving group ("backside attack").  In this case, one of the carbonyl $\ce{C-O}$ bonds is the leaving group.

Using simple trigonometry we can estimate what the angle is for 180° attack of the $\ce{C-O}$ bond relative to the plane of the carbon and oxygen nuclei. Half of the $\ce{O-C-O}$ bond angle is 51.4°, leaving 38.6° to reach 90°. So the angle for backside attack relative to the plane containing the $\ce{C}$ and $\ce{O}$ atoms is $\mathrm{90°+38.6°=128.6°}$ 

The actual angle for attack has been found to be around 107° (see @Philipp 's comment on the Bürgi-Dunitz angle).  This angle was first determined by extensive, detailed crystallographic investigations and later supported by computational studies. The approach used here is no more than a quick, "back of the envelope" method. Yet it still provides an answer within about 20° of Bürgi and Dunitz's work. For such a quick, crude method, I would consider this result to be in qualitative agreement. It supports the back side attack $\mathrm{SN2}$ model, illustrates how a $\ce{C-O}$ bond participates as a leaving group and how we wind up with a tetrahedral carbon.