Question

半径为R的圆外接与三角形ABC 且2R(sin^2A-sin^2c)=(根号3*a-b)sinB求角C

Answer 1

2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB

有正弦定理

2RsinA=a,2RsinC=c

所以(a+c)(sinA-sinC)=(√3a-b)sinB

sinA=a/2R,sinB=b/2R,sinC=c/2R

所以(a+c)(a-c)=(√3a-b)b

a^2-c^2=√3ab-b^2

a^2+b^2-c^2=√3ab

cosC=(a^2+b^2-c^2)/2ab=√3/2

C=30度