Question

在三角形ABC中,角A,B,C的对边分别是a,b,c,已知M=(SinC,CosC),N=(1,-根号3).C=2,MN？

Answer 1

MN=0
则M垂直N,
(sinC*1+cosC*(-√3)=0,
sinC*1/2-√3/2*cosC=0,
sin(C-60)=0,
C=60度.
A+B=180-C,(A+B)/2=90-(C/2).
sin(A+B)/2=cos(C/2)=cos45.
Sin2〔(A+B)/2〕+Cos2C
=(cos45)^2+cos120
=1/2-1/2
=0.
2.若SinC+Sin(B-A)=2Sin2A
2sin(C+B-A)/2*cos(C+A-B)/2=2Sin2A,
而,2sin(C+B-A)/2=cosA,cos(C+A-B)/2=sinB,
即有,cosA*sinB=sin2A=2sinA*cosA,
cosA(sinB-2sinA)=0,
cosA=0,sinB-2sinA=0,
A=90度,
sinB=2sinA,
b=2a,而,A=90度,a^2=b^2+c^2,c=2.
a^2=4a^2+4,(不合,舍去).
A≠90度,则有
b=2a,
而,C=60度,
cos60=(a^2+b^2-c^2)/2ab,c=2.
1/2=(a^2+4a^2-4)/4a^2,
a^2=4/3,
a=2√3/3.
b=2*2√3/3=4√3/3.
三角形ABC的面积=1/2*sinC*ab=1/2*√3/2*(2√3/3)*(4√3/3)
=2√3/3.,2,在三角形ABC中,角A,B,C的对边分别是a,b,c,已知M=(SinC,CosC),N=(1,-根号3).C=2,MN=0
1,求Sin2〔(A+B)/2〕+Cos2C 2,若SinC+Sin(B-A)=2Sin2A,求三角形ABC的面积