Question

△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

△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②若tanA:tanB:tanC=1:2:3且c=1，求b。

Answer 1

1、因为AB•AC=2√3S，则AB的模乘AC的模乘cosA=2√3S,即bcosA=2√3S
又因为S=1/2bcsinA,则tanA=√3/3 A=30°
2、解题思路，你可以自己来解答
TanA:tanB:tanC=1:2:3化简为sinA/cosA=sinB/cosB=sinB/cosB=1:2:3,

根据正弦定理a/sinA=b/sinB=c/sinC=2R，sinA=a/2R,sinB=b/2R,sinC=c/2R,
代入化简，然后再由余弦定理，将角化为边，联立三个方程组，即可

Answer 2

(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

追问

cot是不是cos啊？

追答

cotA = 1/tanA = cosA/sinA

追问

你第一条用cot我看不懂啊。。

追答

(1)
AB.AC =2√3S
bc .cosA = 2√3S
(1/2)bc sinA .(2cosA/sinA) = 2√3S
2S /tanA = 2√3S
tanA = 1/√3
A =π/6

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