Question

请问一下，这道数学题怎么做？

第二问不要用向量

Answer 1

tanC =-3√3
(1)
tanC=-tan(A+B)
-3√3= -(tanA+tanB)/(1-tanA.tanB)
3√3(1-tanA.tanB) = tanA+tanB
tanA(3√3tanB +1) = 3√3-tanB
tanA =( 3√3-tanB)/(3√3tanB +1)
//
2bcosA=acosB
2bsinA/tanA=asinB/tanB
2(b/sinB)/tanA = (a/sinA)/tanB
2/tanA=1/tanB
2tanB= tanA
2tanB=( 3√3-tanB)/(3√3tanB +1)
2tanB(3√3tanB +1) = 3√3-tanB
6√3(tanB)^2 +3tanB -3√3 =0
tanB = (-3+15)/(12√3) or (-3-15)/(12√3) (rej)
ie
tanB = √3/3

Answer 2

tanC =-3√3
(1)
tanC=-tan(A+B)
-3√3= -(tanA+tanB)/(1-tanA.tanB)
3√3(1-tanA.tanB) = tanA+tanB
tanA(3√3tanB +1) = 3√3-tanB
tanA =( 3√3-tanB)/(3√3tanB +1)
//
2bcosA=acosB
2bsinA/tanA=asinB/tanB
2(b/sinB)/tanA = (a/sinA)/tanB
2/tanA=1/tanB
2tanB= tanA
2tanB=( 3√3-tanB)/(3√3tanB +1)
2tanB(3√3tanB +1) = 3√3-tanB
6√3(tanB)^2 +3tanB -3√3 =0
tanB = (-3+15)/(12√3) or (-3-15)/(12√3) (rej)
ie
tanB = √3/3