14. В треугольнике АВС АС = BC=7, tgA = \(\frac{33}{4\sqrt{33}}\) . Найдите АВ.

14. Дано: \(\triangle ABC\), \(AC = BC = 7\), \(\tan A = \frac{33}{4\sqrt{33}}\) . Найти: \(AB\).

Т.к. \(AC = BC\), то \(\triangle ABC\) - равнобедренный, а углы при основании равны, т.е. \(\angle A = \angle B\).

\(\tan A = \frac{33}{4\sqrt{33}} = \frac{\sqrt{33}}{4}\)

Найдем \(\cos A\):

\(\frac{1}{\cos^2 A} = 1 + \tan^2 A\)

\(\cos^2 A = \frac{1}{1 + \tan^2 A} = \frac{1}{1 + (\frac{\sqrt{33}}{4})^2} = \frac{1}{1 + \frac{33}{16}} = \frac{1}{\frac{16 + 33}{16}} = \frac{16}{49}\)

\(\cos A = \sqrt{\frac{16}{49}} = \frac{4}{7}\)

По теореме косинусов:

\(AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos C\)

Найдем \(\angle C\):

\(\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - 2\angle A\)

\(\cos C = \cos (180^\circ - 2A) = -\cos (2A) = -(\cos^2 A - \sin^2 A)\)

Найдем \(\sin A\):

\(\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - (\frac{4}{7})^2} = \sqrt{1 - \frac{16}{49}} = \sqrt{\frac{49 - 16}{49}} = \sqrt{\frac{33}{49}} = \frac{\sqrt{33}}{7}\)

\(\cos C = -(\cos^2 A - \sin^2 A) = -(\frac{16}{49} - \frac{33}{49}) = -(-\frac{17}{49}) = \frac{17}{49}\)

Тогда:

\(AB^2 = 7^2 + 7^2 - 2 \cdot 7 \cdot 7 \cdot \frac{17}{49} = 49 + 49 - 2 \cdot 49 \cdot \frac{17}{49} = 98 - 2 \cdot 17 = 98 - 34 = 64\)

\(AB = \sqrt{64} = 8\)

Ответ: \(AB = 8\)