Question

△ABC中，角A、B、C的对边分别是a、b、c，已知向量m=(cos3A/2,sin3A/2),n=(cosA/2,sinA/2),

且满足m+n的绝对值=√3
1）求角A
2）若b+c=√3a,试判断△ABC的形状

Answer 1

你好：
（1）。向量m+向量n=(cos3A/2+cosA/2,sin3A/2+sinA/2)
那么|m+n|=√[(cos3A/2+cosA/2)^2+(sin3A/2+sinA/2)^2]
=√[(cos3A/2)^2+(sin3A/2)^2+(cosA/2)^2+(sinA/2)^2+2cos3A/2*cosA/2+2sin3A/2*sinA/2]
=√[2+2cos(3A/2-A/2)]=√(2+2cosA)=√3
去根号得到：2+2cosA=3
cosA=0.5
因为A为三角形的内角，所以A=60°
（2）。因为b+c=√3a,由正弦定理有：sinB+sinC=√3sinA=√3*sin60°=3/2
因为A+B+C=180°,C=180°-B-A
,那么，sinC=sin(180°-B-A )=sin(B+A)=sin(B+60°)=0.5sinB+√3/2cosB
所以：sinB+sinC=sinB+0.5sinB+√3/2cosB=√3(√3/2*sinB+0.5cosB)=√3sin(B+30°)=3/2
所以sin(B+30°)=√3/2
B=30°或B=90°
当B=30°时，C=90°
所以△ABC为直角三角形
回答完毕，谢谢！

Answer 2

(1)
m+n = (cos(3A/2)+cos(A/2),sin(3A/2)+sin(A/2))
|m+n|^2
= (cos(3A/2)+cos(A/2))^2+(sin(3A/2)+sin(A/2))^2
= 2+ 2(cos(3A/2)cos(A/2) + sin3A/2sinA/2)
= 2+ 2cosA
|m+n|=√3
=>2+ 2cosA = 3
cosA = 1/2
A = 60°

(2)
b+c=√3a
(b+c)^2 = 3a^2
b^2+c^2 + 2bc = 3a^2
By cosine-rule
a^2 = b^2+c^2 - 2bccosA
=> b^2+c^2 + 2bc = 3(b^2+c^2 - 2bccosA)
2b^2-5bc-2c^2 =0
(2b-c)(b-2c) =0
b = c/2 or b=2c
case 1:when b=2c
b+c=√3a
3c =√3a
a = √3c

a: b : c = √3 : 2:1
By sine -rule
a/sinA = b/sinB = c/sinC
√3/sin60° = 2/sinB = 1/sinC
1/2 = 2/sinB = 1/sinC
=> B = 90°, C=30°

case 2:when b=c/2
C=90°, B=30°