题目内容
已知△ABC的三个内角∠A,∠B,∠C所对的边分别为a,b,c,且
=-
,则角A的大小为______.
cosA |
cosB |
a |
b+2c |
∵
=-
,∴根据正弦定理,得
=-
,
即sinBcosA+2sinCcosA=-cosBcosA,
整理得-2sinCcosA=sinBcosA+cosBcosA=sin(A+B),
∵在△ABC中,sin(A+B)=sin(π-C)=sinC>0,
∴-2sinCcosA=sinC,约去sinC得cosA=-
.
又∵A∈(0,π),∴A=
.
故答案为:
cosA |
cosB |
a |
b+2c |
cosA |
cosB |
sinA |
sinB+2sinC |
即sinBcosA+2sinCcosA=-cosBcosA,
整理得-2sinCcosA=sinBcosA+cosBcosA=sin(A+B),
∵在△ABC中,sin(A+B)=sin(π-C)=sinC>0,
∴-2sinCcosA=sinC,约去sinC得cosA=-
1 |
2 |
又∵A∈(0,π),∴A=
2π |
3 |
故答案为:
2π |
3 |
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