Question

已知向量M=(a+b,a-c) N=(sinB, sinA-sinB),且M//N，其中ABC是三角形ABC的内角，abc分别是角ABC的对边

求角C（答案+解析）

Answer 1

By sine-rule
a/sinA = b/sinB = c/sinC
=> sinB = bsinC/c and sinA= asinC/c
M//N
=> (a-c)/(a+b) = (sinA-sinB)/(sinB)
(a-c)sinB= (a+b)(sinA-sinB)
(2a+b-c)sinB = (a+b)sinA
(2a+b-c)bsinC/c = (a+b)asinC/c
(2a+b-c)b =(a+b)a
2ab- bc +b^2 = a^2+ab
a^2-b^2 = ab-bc
c = a+b -a^2/b
By cosine rule
c^2 = a^2+b^2 -2abcosC
(a+b-a^2/b)^2 = a^2+b^2 -2abcosC
(a+b)^2- 2(a^2/b)(a+b) +a^4/b^2 = a^2+b^2 -2abcosC
2ab - 2a^3/b- 2a^2 +a^4/b^2 = -2abcosC
cosC = -1 + a^2/b^2 + a/b - a^3/b^3
C = arccos{-1 + a^2/b^2 + a/b - a^3/b^3}