Question

已知锐角三角形ABC ABC所对的边为abc.且2asinB=根号3b.(1)若cosC=5分之4,求sin(A-C)

Answer 1

√3b=2asinB

b/sinB=a/(√3/2)

由正弦定理a/sinA=b/sinB===>sinA=√3/2===>cosA=1/2

sinC=√[1-(4/5)²]=3/5

∴sin(A-C)=(√3/2)(4/5)-(1/2)(3/5)=(4√3-3)/10