В треугольнике abc ac=bc, ab=14, sinA=24/25. найдите AC.

[latex]AC={2AB\cos A}\\\sin^2A+\cos^2A=1\Rightarrow\cos A=\sqrt{1-\sin^2A}\\\cos A=\sqrt{1-\frac{576}{625}}=\sqrt{\frac{49}{625}}=\frac7{25}\\AC=2\cdot14\cdot\frac7{25}=\frac{196}{25}=7\frac{21}{25}=7,84[/latex]