Question

在三角形ABC中,a,b,c分别是角A,B,C的对边,向量m=(b,2a-c),向量n=(cosB,cosC),且向量m平行向量n,则B的大小

答案是3分之π 主要是过程 拜托啦！

Answer 1

过A作AH⊥BC交于H
则a=CH+BH=bcosC+ccosB (1)
∵向量m=(b,2a-c),向量n=(cosB,cosC),且向量m平行向量n
∴b/cosB=(2a-c)/cosC
(2a-c)cosB=bcosC
2acosB=bcoaC+ccosB=a (代入(1))
cosB=1/2
∴B=π/3

Answer 2

By cosine rule
a^2 = b^2+c^2 - 2bc(cosA)

m//n
cosC/cosB = (2a-c)/b
bcosC=(2a-c)cosB
2ab(cosC) = 2a(2a-c)cosB
a^2+b^2-c^2 = 2a(2a-c)cosB
= (4a^2)cosB - 2ac(cosB)
= (4a^2)cosB - a^2-c^2 +b^2
(4a^2)cosB = 2a^2
cosB = 1/2
B = π/3