Question

在△ABC中 a,b,c分别为角A,B,C的对边,若m=(sin(B+C)/2)^2,1)

在△ABC中 a,b,c分别为角A,B,C的对边,若向量m=(sin(B+C)/2)^2,1) ,向量n=(cos2A+7/2,4)，且向量m∥向量n

1）求角A的度数
2）若a=根号3，b+c=3，求△ABC的面积 和 角B的大小

Answer 1

(1)
m//n
=> 1/(sin(B+C)/2)^2 = 4/(cos2A+7/2)
4(sin(B+C)/2)^2 =cos2A+7/2
4(cosA/2)^2 = cos2A+7/2
2(cosA+1) = 2(cosA)^2 - 1+ 7/2
4(cosA)^2-4cosA+1 =0
(2cosA-1)^2 =0
cosA = 1/2
A= π/3
(2)
a=√3，
b+c=3
c = 3-b
a/sinA = b/sinB = c/sinC
√3/sinπ/3 = b/sinB = (3-b)/sin(π-A-B)
2 = b/sinB = (3-b)/sin(2π/3-B)

2 = b/sinB
sinB = b/2 => cosB = √(4-b^2)/2

2 = (3-b)/sin(2π/3-B)
2sin(2π/3-B) = 3-b
2(√3/2cosB+1/2sinB) = 3-b
√3cosB+sinB = 3-b
√3(√(4-b^2)/2) + b/2 = 3-b
√(12-3b^2) = 6-3b
12-3b^2 = 36 -36b+9b^2
b^2-3b+2 =0
(b-1)(b-2) =0
b =1 or 2 (rejected)

b =1 => c=2
△ABC的面积 = 1/2bc sinA
= 1/2(2)(1)(√3/2)
= √3/2
sinB = b/2
= 1/2
B = π/6

Answer 2

向量n=(cos2A+7/2？？？？？？,4)
数值是否有误？求真相！

呵呵，我看错了，一楼已解答。