Question

已知三角形ABC的内角A.B.C的对边分别为a,b,c,且满足sin（2A＋B）/sinA=2＋2cos...

Answer 1

(1)sin(2A+B)/sinA=2+2cos(A+B)
sin(2A+B)=2sinA+2sinAcos(A+B)
sin2AcosB+sinBcos2A=2sinA+2sinA(cosAcosB-sinAsinB)
2sinAcosAcosB+sinBcos2A=2sinA+2sinAcosAcosB-2sinAsinAsinB
sinBcos2A+2sinBsinAsinA=2sinA
有二倍角公式
sinB(cos2A+2sinAsinA)=sinB=2sinA
由正弦定理
b=2a.
(2)
由余弦定理
c*c=7a*a=a*a+b*b-2*a*b*cosC
将b=2a代人俯钉碘固鄢改碉爽冬鲸
7a*a=a*a+4a*a-4a*a*cosC
cosC=-1/2
0<=C<=180
C=150

Answer 2

(1)sin(2a+b)/sina=2+2cos(a+b)
sin(2a+b)=2sina+2sinacos(a+b)
sin2acosb+sinbcos2a=2sina+2sina(cosacosb-sinasinb)
2sinacosacosb+sinbcos2a=2sina+2sinacosacosb-2sinasinasinb
sinbcos2a+2sinbsinasina=2sina
有二倍角公式
sinb(cos2a+2sinasina)=sinb=2sina
由正弦定理
b=2a.
(2)
由余弦定理
c*c=7a*a=a*a+b*b-2*a*b*cosc
将b=2a代人
7a*a=a*a+4a*a-4a*a*cosc
cosc=-1/2
0<=c<=180
c=150