已知三角形ABC中,A>B>C,角B=60,且sinA-sinC+根2/2cos(A-C)=根2/2,求角A,B,C

问题描述:

已知三角形ABC中,A>B>C,角B=60,且sinA-sinC+根2/2cos(A-C)=根2/2,求角A,B,C

∵sinA-sinC+(√2/2)cos(A-C)=√2/2,∴2sin[(A-C)/2]cos[(A+C)/2]+(√2/2)cos(A-C)=√2/2,∴2sin(B/2)sin[(A-C)/2]+(√2/2)-√2{sin[(A-C)/2]}^2=√2/2,而B=60°, ∴...