Question

已知a、b、c是△abc的三边,a、b、c满足关系式(2b)^2=4(c+a)(c-a)且5a-3c=0,求sina+s

Answer 1

(2b)² = 4(c + a)(c - a)

4b² = 4(c² - a²)

b² = c² - a²

a² + b² = c²

所以△ABC 是直角三角形 ,∠C是直角

因为 5a - 3c = 0

所以 a/c = 3/5

所以 b/c = 4/5

sin A = a/c = 3/5

sin B = b/c = 4/5

所以 sinA + sinB = 3/5 + 4/5 = 7/5