已知在三角形ABC中,向量m=(-1,根号3),向量n(cosA,sinA),且向量m×向量n=1.(1)求角A;

问题描述:

已知在三角形ABC中,向量m=(-1,根号3),向量n(cosA,sinA),且向量m×向量n=1.(1)求角A;
(2)若tan(π/4 +B)=-3,求tanc的值?
1个回答 分类:数学 2014-12-16

问题解答:

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m·n
=-cosA+√3sinA=1
-(1/2)cosA+(√3/2)sinA=1/2
cos(2π/3)cosA+sin(2π/3)sinA=1/2
cos(2π/3-A)=cos(π/3)
2π/3-A=π/3
A=π/3
tanB
=tan(π/4 +B -π/4 )
=(tan(π/4 +B)-1)/(tan(π/4 +B)+1)
=(-4)/(-2)
=2
tanA=√3
tanC
=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(2+√3)/(1-2√3)
=(2+√3)(1+2√3)/11
=(8+5√3)/11
 
 
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