Воспользуемся тем, что:
\(1 + {\rm{t}}{{\rm{g}}^2}A = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + {5^2} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \dfrac{1}{{\sqrt {26} }}\).
По определению косинуса из треугольника АВС:
\(\cos A = \dfrac{{AC}}{{AB}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{1}{{\sqrt {26} }} = \dfrac{{AC}}{{13}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,AC = \dfrac{{13}}{{\sqrt {26} }}\).
По определению косинуса из треугольника AНС:
\(\cos A = \dfrac{{AH}}{{AC}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{1}{{\sqrt {26} }} = AH:\dfrac{{13}}{{\sqrt {26} }}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,AH = \dfrac{1}{{\sqrt {26} }} \cdot \dfrac{{13}}{{\sqrt {26} }} = 0,5\).
Тогда: \(BH = AB — AH = 13 — 0,5 = 12,5\).
Ответ: 12,5.