10. В треугольнике АВС АС = ВС = 5, \(\sinA = \frac{7}{25}\). Найдите АВ.

10. Дано: \(\triangle ABC\), \(AC = BC = 5\), \(\sin A = \frac{7}{25}\). Найти: \(AB\).

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

По теореме синусов:

\(\frac{AB}{\sin C} = \frac{AC}{\sin B}\)

Выразим AB:

\(AB = \frac{AC \cdot \sin C}{\sin B} = \frac{AC \cdot \sin C}{\sin A}\)

Найдем \(\cos A\):

\(\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{7}{25})^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{625 - 49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}\)

По теореме косинусов:

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

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

Найдем \(\angle C\):

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

\(\sin C = \sin (180^\circ - 2A) = \sin (2A) = 2 \cdot \sin A \cdot \cos A = 2 \cdot \frac{7}{25} \cdot \frac{24}{25} = \frac{336}{625}\)

\(\cos C = \cos (180^\circ - 2A) = -\cos (2A) = -(\cos^2 A - \sin^2 A) = -(\frac{576}{625} - \frac{49}{625}) = -\frac{527}{625}\)

Тогда:

\(AB = \sqrt{5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot (-\frac{527}{625})} = \sqrt{25 + 25 + \frac{26350}{625}} = \sqrt{50 + \frac{1054}{25}} = \sqrt{\frac{1250 + 1054}{25}} = \sqrt{\frac{2304}{25}} = \frac{48}{5} = 9.6\)

Ответ: \(AB = 9.6\)