Question

高三数学题第20题

Answer 1

按余弦定理: cosC = (a² + b² - c²)/(2ab)
代入得: 3a - 3b(a² + b² - c²)/(2ab) = √3csinB
√3csinB = 3a - 3(a² + b² - c²)/(2a) = 3(2a² - a² - b² + c²)/(2a) = 3(a² + c² - b²)/(2a)
√3sinB = 3(a² + c² - b²)/(2ac) = 3cosB
sinB/cosB = √3 = tanB
B = 60°

b/sinB = 2/sin60° = a/sinA = c/sinC
a = 2sinA/sin60° = 4sinA/√3, c = 4sinC/√3
y = ac = (16/3)sinAsinC = (8/3)[cos(A - C) - cos(A + C)]
= (8/3)[cos(A - C) - cos120°]
A = C时, cos(A - C) = 1， y最大，为(8/3)(1 + 1/2) = 4