△ABC中,角A,B,C所对应的边分别为a,b,c面积为S,①若向量AB•AC=2√3S,求A的值
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解决时间 2021-02-10 04:52
- 提问者网友:你在我心中是最美
- 2021-02-09 06:02
△ABC中,角A,B,C所对应的边分别为a,b,c面积为S,①若向量AB•AC=2√3S,求A的值
最佳答案
- 二级知识专家网友:如果这是命
- 2021-02-09 06:52
(1)
AB.AC =2√3S
bc .cosA = 2√3S
(1/2)bc sinA .(2cotA) = 2√3S
2S . cotA = 2√3S
cotA = √3
A =π/6
(2)
c=1
tanA:tanB:tanC=1:2:3
=> tanA = k , tanB = 2k, tanC =3k
tanC = -tan(A+B)
= -(tanA+tanB)/(1- tanA.tanB)
3k = -(3k)/(1-2k^2)
1-2k^2 =-1
k= 1
tanB = 2
sinB = 2/√5
tanC = 3
sinC = 3/√10
c/sinC = b/sinB
1/( 3/√10) = b/(2/√5)
b = (√10/3)(2/√5)
= √2/6
AB.AC =2√3S
bc .cosA = 2√3S
(1/2)bc sinA .(2cotA) = 2√3S
2S . cotA = 2√3S
cotA = √3
A =π/6
(2)
c=1
tanA:tanB:tanC=1:2:3
=> tanA = k , tanB = 2k, tanC =3k
tanC = -tan(A+B)
= -(tanA+tanB)/(1- tanA.tanB)
3k = -(3k)/(1-2k^2)
1-2k^2 =-1
k= 1
tanB = 2
sinB = 2/√5
tanC = 3
sinC = 3/√10
c/sinC = b/sinB
1/( 3/√10) = b/(2/√5)
b = (√10/3)(2/√5)
= √2/6
全部回答
- 1楼网友:丢不掉的轻狂
- 2021-02-09 07:36
解:因为ab向量×ac向量=(8/3)*s△abc
所以c*b*cosa=(8/3)*(1/2)*b*c*sina
即cosa=(4/3)*sina
则tana=sina/cosa=3/4
可知角a是锐角,则由sina²+cos²a=1可得:
sina=3/5,cosa=4/5
(1)因为b+c=180°-a,且cos2a=1-2sin²a=1-2*(9/25)=7/25
所以:sin²[(b+c)/2]+cos2a
=sin²(90°-a/2)+(7/25)
=cos²(a/2)+(7/25)
=(1+cosa)/2 +(7/25)
=(9/10)+(7/25)
=59/50
(2)若b=2,s△abc=(1/2)*b*c*sina=3,则:
(1/2)*2*c*(3/5)=3
解得c=5
则由余弦定理可得:
a²=b²+c²-2bc*cosa
=4+25-2*2*5*(4/5)
=13
所以解得a=√13
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