(1)根据正弦定理,有:
sinA - sinB=sinBcosC
sin[π-(B+C)] - sinB - sinBcosC
sin(B+C) - sinB - sinBcosC=0
sinBcosC + sinCcosB - sinB - sinBcosC=0
则sinCcosB=sinB
∵B是△ABC的内角
∴sinB≠0
∴两边同除以sinB,得:
sinCcosB/sinB=1
sinC•cotB=1
则sinC/tanB=1
追答(2)由已知:1 - 2=2cosC
∴cosC=-1/2
根据余弦定理:c²=a²+b²-2abcosC
=1²+2²-2•1•2•(-1/2)
=1+4+2=7
∴c=√7
追问第一步后面cot是余切?没学过啊
能换成其他吗
追答好的,第(1)倒数第二行改一下:
sinC/(sinB/cosB)=1
就是把原先在分子上的cosB换成分母上
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