Воспользуемся тем, что:
\(1 + {\rm{t}}{{\rm{g}}^2}A = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{{{33}^2}}}{{16 \cdot 33}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{33}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\frac{{49}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \frac{4}{7}\).
По определению синуса и косинуса из треугольника АВС:
\(\cos A = \frac{{AC}}{{AB}};\,\,\,\,\,\,\,\,\,\,\sin B = \frac{{AC}}{{AB}}\).
Следовательно, \(\sin B = \cos A = \frac{4}{7}\).
По определению синуса из треугольника ВCH:
\(\sin B = \frac{{CH}}{{CB}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{4}{7} = \frac{{CH}}{7}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,CH = 4\).
Ответ: 4.