Воспользуемся тем, что:
\(1 + {\rm{t}}{{\rm{g}}^2}A = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{{{33}^2}}}{{16 \cdot 33}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\,1 + \frac{{33}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\frac{{49}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\cos A = \frac{4}{7}\).
По определению косинуса из треугольника ABC:
\(\cos A = \frac{{AC}}{{AB}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{4}{7} = \frac{4}{{AB}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,AB = 7\).
Ответ: 7.