11. В треугольнике АВС АС = BC, AB = 9,6, \(\sinA = \frac{7}{25}\). Найдите АС.

11. Дано: \(\triangle ABC\), \(AC = BC\), \(AB = 9.6\), \(\sin A = \frac{7}{25}\). Найти: \(AC\).

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

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:

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

Т.к. \(AC = BC\), то

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

\(AB^2 = 2AC^2(1 - \cos C)\)

Найдем \(\angle C\):

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

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

Тогда:

\(AB^2 = 2AC^2(1 - (-\frac{527}{625})) = 2AC^2(1 + \frac{527}{625}) = 2AC^2(\frac{625 + 527}{625}) = 2AC^2(\frac{1152}{625})\)

\(AC^2 = \frac{AB^2}{2} \cdot \frac{625}{1152}\)

\(AC = \sqrt{\frac{AB^2}{2} \cdot \frac{625}{1152}} = AB \cdot \sqrt{\frac{625}{2304}} = AB \cdot \frac{25}{48} = 9.6 \cdot \frac{25}{48} = 0.2 \cdot 25 = 5\)

Ответ: \(AC = 5\)