在三角形ABC中,已知b^2 +c^2 =a^2 +根号3bc.求∠A的大小求2sinBcosC-sin(B-C)的值

问题描述:

在三角形ABC中,已知b^2 +c^2 =a^2 +根号3bc.求∠A的大小
求2sinBcosC-sin(B-C)的值

由余弦定理:CosA=(c²+b²-a²)/2bc ,
∵ b²+c²=a²+√3*bc ,
∴ CosA=(a²+√3*bc-a²)/2bc =√3/2 ,
∴ ∠A=30°;
∴2sinBcosC-sin(B-C)
= 2sinBcosC-(sinBcosC-cosBsinC)
= 2sinBcosC-sinBcosC+cosBsinC
= sinBcosC+cosBsinC
= sin(B+C)
= sinA
= sin30°
= 1/2 。

1.
b^2 +c^2 =a^2 +√3bc
a^2=b^2 +c^2 -√3bc
a^2=b^2 +c^2-2bccosA
两式相减得
2bccosA=√3bc
cosA=√3/2
A=π/6
2.
2sinBcosC-sin(B-C)
=2sinBcosC-sinBcosC+sinCcosB
=sinBcosC+sinCcosB
=sin(B+C)
=sinA
=1/2

答案:∠A=30°
上式=1/2
∵b^2 +c^2 =a^2+√3bc
∴b^2+c^2-a^2=√3bc
∴cosA=(b^2+c^2-a^2)/2bc=√3bc/2bc=√3/2
∴A=30°
2sinBcos-sin(B-C)
=2sinBcosC-sinBcosC+sinCcosB
=sinBcosC+sinCcosB
=sin(B+C)
=sinA
=1/2
呼呼,累死了.厄,懂了吗?

答案:30度 1/2
解析:1.cosA=b^2 +c^2-a^2/2bc
因为b^2 +c^2 =a^2 +根号3bc 即 b^2 +c^2-a^2=根号3bc
所以 cosA=根号3bc /2bc=根号3/2 所以角A=30
2.因为 sin(B-C)=sinBcosC-sinccosb
所以 2sinBcosC-sin(B-C)=2sinBcosC-sinBcosC+sinccosb=sinBcosC+sinccosb=sin(B+C)=sin(180-A)=sinA=1/2