Question

已知锐角三角形ABC中，sin（A＋B）＝3／5，sin（A－B）＝1／5．＊设AB＝3，求AB边上的高．

答案是根号6,谁算给我看?

Answer 1

设高与AB交于D
2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)-sin(A-B)
sinAcosB=2/5
cosAsinB=1/5
A,B,C都为锐角A+B>90
cos(A+B)=-4/5
2cosAcosB=cos(A+B)-sin(A-B)
cosAcosB=-1/2
tgA=(2/5)/(-1/2)=CD/AD
tgB=(1/5)/(-1/2)=CD/BD
AD/CD+BD/CD=(AD+BD)/CD=AB/CD=3/CD
=>CD