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

sinA=a/(2R)=a/(2√2),sinC=c/(2√2),sinB=b/(2√2)

2√2(sinA^2-sinC^2)=(a-b)SinB

2√2(a^2/8-c^2/8)=(a-b)*b/(2√2)

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

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

∠C=60°