Треугольник АВС равнобедренный, поэтому \(\angle \,CAB = \angle \,HBA\).
Следовательно, \({\rm{tg}}\,B = \frac{{33}}{{4\sqrt {33} }}\).
Воспользуемся тем, что:
\(1 + {\rm{t}}{{\rm{g}}^2}B = \frac{1}{{{{\cos }^2}B}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{{{33}^2}}}{{16 \cdot 33}} = \frac{1}{{{{\cos }^2}B}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{33}}{{16}} = \frac{1}{{{{\cos }^2}B}}\,\,\,\,\,\,\,\, \Leftrightarrow \)
\(\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\frac{1}{{{{\cos }^2}B}} = \frac{{49}}{{16}}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos B = \frac{4}{7}\).
По определению косинуса из треугольника ABH:
\(\cos B = \frac{{BH}}{{AB}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\frac{4}{7} = \frac{{BH}}{7}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,BH = 4\).
Ответ: 4.