在三角形ABC中,b=asinC,c=acosB,判断三角行ABC的形状

显然sinC≤1,cosB≤1,所以b≤a,c≤a由a/sinA=b/sinB=c/sinC得sinB=sinAsinC,sinC=sinAcosB,所以(sinB)^2=(sinAsinC)^2,(sinC)^2=(sinA)^2(1-(sinB)^2)所以(sinB)^2=(sinA)^2/[1+(sinA)^2],(sinC)^2=1/[1+(sinA)^2]所...