以B为原点,C(1,0)
由等腰三角形ABC,得出A(cos80, sin80)
由正弦定理,CD / sin60 = BC / sin40,
得出CD=sin60 / sin40,
得出D(1- cos80.sin60 / sin40, sin80.sin60 / sin40)
AD高度差=sin80.sin60 / sin40 - sin80 = 2.sin80/sin40.cos50.sin10=sin20
AD横向差= 1- cos80.sin60 / sin40 - cos80
=1-cos80(sin60 / sin40+1)
=1-2.cos80/sin40.sin50.cos10
=1-2.sin10/sin40.sin50.cos10
=1-sin20/sin40.sin50
=1-sin20*cos40/sin40
=(sin40-sin20.cos40)/sin40
AD横向差/AD高度差 =(sin40-sin20.cos40)/sin40/sin20
=(2.cos20-cos40)/sin40
=(2.cos(60-40)-cos40)/sin40
=(2.cos60.cos40+2.sin60.sin40-cos40)/sin40
=2.sin60.sin40/sin40
=3^0.5
所以AD倾角为30度,角DAB为130度,角ADB为30度。