Question

在△ABC中,2√2(sinA^2-sinC^2)=(a-b)SinB,△ABC外接圆半径为√2,求角C的度数.

Answer 1

△ABC外接圆半径为√2

R=√2

由正弦定理得

a=2RsinA

sinA=a/2√2

sin^2 A=a^2/8

sin^2 C=c^2/8

sinB=b/2√2

2√2(sinA^2-sinC^2)=(a-b)SinB

2√2(a^2-c^2)/8=(a-b)b/2√2

a^2-c^2=ab-b^2

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

由余弦定理得

cosC=(a^2+b^2-c^2)/2ab=1/2

C=60