По определению синуса из треугольника АВС:
\(\sin A = \dfrac{{BC}}{{AB}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{1}{4} = \dfrac{{BC}}{{4\sqrt {15} }}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,BC = \sqrt {15} \).
\(\sin A = \dfrac{{BC}}{{AB}};\,\,\,\,\,\,\,\,\,\,\cos B = \dfrac{{BC}}{{AB}}\).
Следовательно, \(\cos B = \sin A = \dfrac{1}{4}\).
По определению косинуса из треугольника ВНС:
\(\cos B = \dfrac{{BH}}{{BC}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{1}{4} = \dfrac{{BH}}{{\sqrt {15} }}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,BH = \dfrac{{\sqrt {15} }}{4}\).
По теореме Пифагора из треугольника ВНС:
\(B{C^2} = B{H^2} + C{H^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,C{H^2} = {\left( {\sqrt {15} } \right)^2} — {\left( {\dfrac{{\sqrt {15} }}{4}} \right)^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,CH = 3,75\).
Ответ: 3,75.