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