Question

在三角形ABC中，a,b,c分别是角A，B，C的对边，若（a+b+c）(sinA+sinB-sinC)=3asinB,则A+B=？

Answer 1

(sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB
(sinA+sinB)²-sin²C=3sinAsinB
sin²A+2sinAsinB+sin²B-sin²(A+B)=3sinAsinB
sin²A+sin²B-(sinAcosB+cosAsinB)²=sinAsinB
sin²A+sin²B-sin²Acos²B-2sinAcosBcosAsinB-cos²Asin²B=sinAsinB
2sin²Asin²B-2sinAcosBsinBcosA=sinAsinB
cosAcosB-sinAsinB=-1/2
cos(A+B)=-1/2
A+B=2π/3
所以C=π-(A+B)=π/3
S三角形ABC=1/2absinC
1/2ab×sin(π/3)=10√3
ab=40(1)
根据题意
a+b+c=20则a+b=20-c
余弦定理
c²=a²+b²-2abcosC
c²=a²+b²+2ab-3ab
c²=(a+b)²-3ab
c²=(20-c)²-120
c²=400-40c+c²-120
40c=280
c=7

Answer 2

（a+b+c）(sinA+sinB-sinC)=3asinB
正弦定理 (a+b+c)(a+b-c)=3ab
(a+b)^2-c^2=3ab
a^2+b^2-c^2=ab
(a^2+b^2-c^2)/2ab=cosC=1/2
A+B=120度