Question

若a,b,c分别表示三角形ABC内角ABC所对的边长，且（sinA+sinB+sinC)(sinA+sinB-sinC)=3sinAsinB.若三角形面

为10根号3，周长20，求边c

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