Задача 13. В треугольнике ABC \(AC = BC = 7,\;\;{\text{tg}}\,A = \dfrac{{4\sqrt {33} }}{{33}}.\) Найдите высоту CH.
ОТВЕТ: 4.
\(1 + {\rm{t}}{{\rm{g}}^2}A = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \dfrac{{16 \cdot 33}}{{{{33}^2}}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \dfrac{{16}}{{33}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\dfrac{{49}}{{33}} = \dfrac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \dfrac{{\sqrt {33} }}{7}\). По основному тригонометрическому тождеству: \({\sin ^2}A + {\cos ^2}A = 1\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{\sin ^2}A = 1 — \dfrac{{33}}{{49}}\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin A = \dfrac{4}{7}\). По определению синуса из треугольника ACH: \(\sin A = \dfrac{{CH}}{{AC}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\dfrac{4}{7} = \dfrac{{CH}}{7}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,CH = 4\). Ответ: 4.
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