We can write \begin{align*} \tan \left( \frac{B - C}{2} \right) \tan \frac{A}{2} &= \frac{\sin (\frac{B - C}{2}) \sin \frac{A}{2}}{\cos (\frac{B - C}{2}) \cos \frac{A}{2}} \\ &= \frac{\cos (\frac{A + C - B}{2}) - \cos (\frac{A + B - C}{2})}{\cos (\frac{A + B - C}{2}) + \cos (\frac{A + C - B}{2})} \\ &= \frac{\cos (90^\circ - B) - \cos (90^\circ - C)}{\cos (90^\circ - C) + \cos (90^\circ - B)} \\ &= \frac{\sin B - \sin C}{\sin C + \sin B}.\end{align*} As usual, let $ a = BC, $ $ b = AC, $ and $ c = AB $. By the Law of Sines, $ \frac{b}{\sin B} = \frac{c}{\sin C}, $ so \[\frac{\sin B - \sin C}{\sin C + \sin B} = \frac{b - c}{b + c} = \frac{1}{29}.\]Then $ 29b - 29c = b + c, $ so $ 28b = 30c, $ or $ \frac{b}{15} = \frac{c}{14} $. Similarly, we can show that \[\tan \left( \frac{C - A}{2} \right) \tan \frac{B}{2} = \frac{c - a}{c + a},\]so $ \frac{c - a}{c + a} = \frac{1}{27} $. Then $ 27c - 27a = c + a, $ so $ 26c = 28a, $ or $ \frac{a}{13} = \frac{c}{14} $. Finally, \[\tan \left( \frac{A - B}{2} \right) \tan \frac{C}{2} = \frac{a - b}{a + b} = \frac{13 - 15}{13 + 15} = \frac{-2}{28} = \boxed{-\frac{1}{14}}.\]