Question

在锐角三角形ABC中,已知内角A,B,C所对的边分别为a,b,c,向量m=(2sin(A+C),根号3),n=（cos2B,2cos平方B/2-1）

且向量m,n共线
（1）求角B的大小；（2）如果b=1，求三角形ABC的面积最大

Answer 1

(1)
m,n共线
=>
2sin(A+C)/√3 = cos2B/(2(cos(B/2))^2-1)
2sin(A+C) . cosB = √3cos2B
2sinBcosB = √3cos2B
sin2B=√3cos2B
tan2B = √3
B = π/6
(2)
b=1
by sine-rule
a/sinA =b/sinB
a = (b/sinB)sinA
=2sinA
and
c= 2sinC = 2sin(5π/6-A)
A1=area of ABC
=(1/2)acsinB
= (1/2)(2sinA)(2sin(5π/6-A))(1/2)
= sinAsin(5π/6-A)
=(1/2)(cos(2A-5π/6) - cos5π/6)
=(1/2)(cos(2A-5π/6) +√3/2)
max A1 at cos(2A-5π/6)=1
max A1 = (1/2)(1+√3/2)