Question

已知三角形abc中,2√2×（sina×sina-sinC×sinC）＝(a-b）sinb ,三角形abc的外接圆半径为

Answer 1

(1)

利用正弦定理和余弦定理

a/sinA=b/sinB=c/sinc=2R=2√2

a^2+b^2-c^2=2ab*cosC

及2√2（sinA^2-sinC^2）=（a-b）sinB

得cosC=1/2

C=60°

(2)

S=1/2*ab*sinc

=2√3*sinA*sinB

=-√3(cos(A+B)-cos(A-B))

=-√3(cos120°-cos(A-B))

=-√3(-1/2-cos(A-B))

=√3(1/2+cos(A-B))

≤3√3/2

当且仅当A=B=C=60°时,三角形ABC的面积最大值 为3√3/2