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