Воспользуемся тем, что:
\(1 + {\rm{t}}{{\rm{g}}^2}A = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \dfrac{{16 \cdot 33}}{{{{33}^2}}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \dfrac{{16}}{{33}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{{49}}{{33}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \dfrac{{\sqrt {33} }}{7}\).
По определению синуса и косинуса из треугольника АВС:
\(\cos A = \dfrac{{AC}}{{AB}};\,\,\,\,\,\,\,\,\,\,\sin B = \dfrac{{AC}}{{AB}}\).
Следовательно, \(\sin B = \cos A = \dfrac{{\sqrt {33} }}{7}\).
По определению синуса из треугольника ВCH:
\(\sin B = \dfrac{{CH}}{{CB}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{{\sqrt {33} }}{7} = \dfrac{{CH}}{7}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,CH = \sqrt {33} \).
По теореме Пифагора из треугольника BCH:
\(B{C^2} = B{H^2} + C{H^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,B{H^2} = {7^2} — {\left( {\sqrt {33} } \right)^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,BH = 4\).
Ответ: 4.