Question

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

Answer 1

$AC={2ABcos A}\sin^2A+cos^2A=1Rightarrowcos A=sqrt{1-sin^2A}\cos A=sqrt{1-frac{576}{625}}=sqrt{frac{49}{625}}=frac7{25}\AC=2cdot14cdotfrac7{25}=frac{196}{25}=7frac{21}{25}=7,84$