三角形DBE
(1) DB/sinX=BE/sin(180-x-20)=BE/sin(x+20)
三角形EBC
(2) BC/sin40=BE/sin80
(2)/(1) and DB=BC =>
sinX/sin40=sin(X+20)/sin80
sinX/sin40=sin(X+20)/2sin40cos40
sinX=sin(X+20)/2cos40
=> 2cos40*sinX=sin(X+20)
so X=30
三角形DBE
(1) DB/sinX=BE/sin(180-x-20)=BE/sin(x+20)
三角形EBC
(2) BC/sin40=BE/sin80
(2)/(1) and DB=BC =>
sinX/sin40=sin(X+20)/sin80
sinX/sin40=sin(X+20)/2sin40cos40
sinX=sin(X+20)/2cos40
=> 2cos40*sinX=sin(X+20)
so X=30
• 要求不用三角公式而用初等几何的方法解题. -redplum- ♀ (131 bytes) () 11/16/2011 postreply 19:27:35