Question

已知锐角三角形ABC中,sin(A+B)=3/5,sin(A-B)=1/5.(1证明tanA=2tanB(2)设AB=3

Answer 1

(1)sin(A＋B)/sin(A-B)=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)=3,

(tanA+tanB)/(tanA-tanB)=3,

tanA+tanB=3(tanA-tanB)

tanA=2tanB

(2)cos(A+B)=4/5,cos(A-B)=2√6/5.

sinAsinB=(cos(A-B)-cos(A+B))/2=(2-√6)/5

h/tanA+h/tanB=3

h=3(tanA*tanB)/(tanA+tanB)

=3sinAsinB/sin(A+B)=2-√6