Задача 36. В треугольнике ABC угол C равен \({90^ \circ }\), \(AB = 13,\;\;{\text{tg}}\,A = \frac{1}{5}\). Найдите высоту CH.
\(1 + {\rm{t}}{{\rm{g}}^2}A = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{1}{{25}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \frac{5}{{\sqrt {26} }}\). По определению косинуса из треугольника АВС: \(\cos A = \frac{{AC}}{{AB}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{5}{{\sqrt {26} }} = \frac{{AC}}{{13}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,AC = \frac{{5 \cdot 13}}{{\sqrt {26} }}\). По определению косинуса из треугольника AНС: \(\cos A = \frac{{AH}}{{AC}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{5}{{\sqrt {26} }} = AH:\frac{{5 \cdot 13}}{{\sqrt {26} }}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,AH = \frac{{5 \cdot 5 \cdot 13}}{{{{\left( {\sqrt {26} } \right)}^2}}} = \frac{{25 \cdot 13}}{{26}} = \frac{{25}}{2} = 12,5\). По теореме Пифагора из треугольника АНС: \(A{C^2} = A{H^2} + C{H^2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,C{H^2} = {\left( {\frac{{5 \cdot 13}}{{\sqrt {26} }}} \right)^2} — {\left( {\frac{{25}}{2}} \right)^2}\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\,CH = \sqrt {\frac{{25 \cdot 13}}{2} — \frac{{625}}{4}} = \sqrt {\frac{{650 — 625}}{4}} = 2,5\) Ответ: 2,5.
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