已知a,b,c分别为△ABC的内角A,B,C的对边,向量m=（根号3sinA,sinB),向量n=(cosB,根号3cosA),且m·n=1+cos(A+B)1)求角C的大小2）若a+b=4,c=2根号3,求△ABC的面积

Question

已知a,b,c分别为△ABC的内角A,B,C的对边,向量m=（根号3sinA,sinB),向量n=(cosB,根号3cosA),且m·n=1+cos(A+B)
1)求角C的大小
2）若a+b=4,c=2根号3,求△ABC的面积

Answer 1

(1)向量m·向量n = √3(sinAcosB + cosAsinB) = √3sin(A+B) = √3sinC,而cos(A+B) = -cosC
∵m·n=1+cos(A+B) ,∴√3sinC + cosC = 1 = 2·sin[C + (π/6)] ,∴sin[C+(π/6)] = 1/2 ,
而[C + π/6]∈(π/6 ,7π/6) ,∴C + π/6 = 5π/6 ,∴C = 2π/3
(2)根据余弦定理：
c^2 = a^2 + b^2 - 2ab·cosC = a^2 + b^2 + ab = (a + b)^2 - ab
∴ab = 4^2 - 12 = 4
∴S△ABC = (1/2)·ab·sinC = √3