Задача 50. В треугольнике ABC угол C равен \({90^ \circ }\), CH — высота, \(AC = 7,\;{\text{tg}}\,A = \frac{{33}}{{4\sqrt {33} }}.\) Найдите AH.

Ответ

ОТВЕТ: 4.

Решение

Воспользуемся тем, что:

\(1 + {\rm{t}}{{\rm{g}}^2}A = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{{{33}^2}}}{{16 \cdot 33}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,1 + \frac{{33}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\,\frac{{49}}{{16}} = \frac{1}{{{{\cos }^2}A}}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\cos A = \frac{4}{7}\).

По определению косинуса из треугольника ACH:

\(\cos A = \frac{{AH}}{{AC}}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\frac{4}{7} = \frac{{AH}}{7}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,AH = 4\).

Ответ:  4.