如右图所示,CD是平面镜,光线从A点出发经CD上点E反射后照到B点,AC⊥CD,BD⊥CD,垂足分别为C.D,且AC=3,
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设法线为EF(过E点 与CD垂直的线)
AC//EF//BD
角EAC = α = 角EBD
所以 Rt△ACE 相似于 Rt△BDE
EC/ED = AC/BD = 3/6 = 1/2
EC+ED = CD = 11
EC = 11*(1/3) = 11/3
ED = 11*(2/3) = 22/3
Rt△ACE 中
tanα = CE/AC = (11/3)/3 = 11/9
AC//EF//BD
角EAC = α = 角EBD
所以 Rt△ACE 相似于 Rt△BDE
EC/ED = AC/BD = 3/6 = 1/2
EC+ED = CD = 11
EC = 11*(1/3) = 11/3
ED = 11*(2/3) = 22/3
Rt△ACE 中
tanα = CE/AC = (11/3)/3 = 11/9
参考资料: 数学问题请教!!!!!_百度知道 http://zhidao.baidu.com/question/19033070.html
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