已知銳角三角形ABC中,sin(A B)=3/5,sin(A-B)=1/5.(1証明tanA=2tanB(2)設AB=3,求AB邊上的高
要解題過程網友廻答：
- (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
admin=35,sin(A-B)=15.(1証明tanA=2tanB(2)設AB=3,求AB邊上的高要解題過程.jpg "已知銳角三角形ABC中,sin(A+B)=35,sin(A-B)=15.(1証明tanA=2tanB(2)設AB=3,求AB邊上的高要解題過程,第1張")
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