Question

已知三角形ABC的外接圆半径为R,且满足2R(sin方A-sin方C)=(根号3a-b)sinB,求角C

三角形面积的最小值

Answer 1

正弦定理
a/sinA=b/sinB=c/simC=2R
a=2RsinA b=2RsinB c=2RsinC
2R(sin方A-sin方C)=(根号3a-b)sinB
4r^2(sin方A-sin方C)=(根号3a-b)sinB*2R
a^2-c^2=√3ab-b^2
a^2+b^2-c^2=√3ab
余弦定理
cosC=(a^2+b^2-c^2)/2ab=√3ab/2ab=√3/2 C=30°

Answer 2

2. a/sinA=b/sinB=c/sinC=2R=2√ =>a=2RsinA,b=2RsinB,c=2RsinC C）=2R（a-b）sinB =>a 2;-c 2;=(a-b)b =>(a 2;+b 2;-