Question

数学题求各位大大解一下

Answer 1

解：设直线EC与BD相交于O，令DB=DC=x
根据正弦定理：△BCD中，x/sin40=BC/sin100
BC=(sin100/sin40)x
△BEC中，BC/sin50=EC/sin110
EC=(sin110/sin50)*(sin100/sin40)x
根据余弦定理：△EDC中，DE^2=DC^2+EC^2-2DC*EC*cos20
DE^2=x^2+[(sin110/sin50)*(sin100/sin40)x]^2-2(sin110/sin50)*(sin100/sin40)x^2*cos20
根据正弦定理，DE/sin20=EC/sin∠EDC
sin∠EDC=(EC/DE)*sin20
=[(sin110/sin50)*(sin100/sin40)x]/√{x^2+[(sin110/sin50)*(sin100/sin40)x]^2-2(sin110/sin50)*(sin100/sin40)x^2*cos20}*sin20
=(sin110sin100sin20/sin50sin40)/√[1+(sin110sin100/sin50sin40)^2-2(sin110sin100/sin50sin40)cos20]
=(sin110sin100sin20)/√[(sin50sin40)^2+(sin110sin100)^2-2sin110sin100sin50sin40cos20]
=(cos20cos10sin20)/√[(cos40sin40)^2+(cos20cos10)^2-2cos20cos10cos40sin40cos20]
=(sin40cos10)/√[(2cos40sin40)^2+(2cos20cos10)^2-4cos20cos10sin80cos20]
=(sin40cos10)/sin80
=sin40
因为∠EDC>∠BDC=100°
所以∠EDC=180°-40°=140°

Answer 2

110度啦，很简单的相似三角形嘛

追问

不对吧 BDC都100度拉 有过程么？