Playing Billiards

Python 2.7, 352 344 281 bytes

from math import*
def l(a,b,n):
 a*=1.;b*=1.
 r=set()
 for i in range(1,n+3):
  t=[]
  for k in range(1,i):
   for h in[0,.5]:
    x=(i-k-h)
    if 1-(x/k in r):r.add(x/k);t+=(x*a,k*b),
 d=(a*n+1)**2+(b*n+1)**2
 for x,y in t:
  if x*x+y*y<d:d=x*x+y*y;o=degrees(atan(y/x))
 return o

• -16 bytes thanks to @Dschoni

Explanation: instead calculating the cushions hits, I'm adding n tables and taking the new holes as valid : Black border/holes is the original, green border/holes is the valid for n=1, red border/holes is the valid for n=2 and so on. Then I remove the invalid holes (e.g. the blue arrow for n=1). I'll have a list of valid holes and their coordinates, then I calculate their distance from initial point, and then the angle of the smaller distance.
Notes:
a=4.76, b=3.64, n=27 - give 52.66286, trying to figure out why fixed, and saved 8 bytes in the process =D
a=43, b=21, n=10005 - takes ~80 seconds (but gives the right angle)

readable version:

from math import *
def bill(a,b,n):
    a=float(a)
    b=float(b)
    ratios = set()
    for i in range(0,n+2): # Create the new boards
        outter = []
        j=i+1
        for k in range(1,j): # Calculate the new holes for each board
            #y=k
            for hole_offset in [0,0.5]:
                x=(j-k-hole_offset)
                if (x/k) not in ratios:
                    ratios.add(x/k)
                    outter.append((x*a,k*b))
    min_dist = (a*n+1)**2+(b*n+1)**2
    for x,y in outter:
        if x*x+y*y<min_dist:
            min_dist = x*x+y*y
            min_alpha=degrees(atan(y/x))
    return min_alpha

Haskell, 133 117 bytes

This is my implementation:

With a 2x1 table, a path will hit exactly n cushions before going into a pocket if: (x-1)/2 + (y-1) == n and x,y are mutually primes. where x,y are the distance of the ball over horizontal/vertical axes.

Paths are the same with arbitrary table size, so we just have to update lengths and angles with (a,b) and keep the shortest. Path length is sqrt((x*a/2)^2+(y*b)^2) and angle is atan((y*b)/(x*a/2))

z=toEnum
f a b n=minimum[[z x^2+r^2,180/pi*atan(r/z x)]|x<-[1..2*n+2],y<-[n+1-div(x-1)2],r<-[2*b/a*z y],gcd x y<2]!!1