设AD=1
AB=1,BD=2sin50,AC=2sin50,BC=2sin50-1
CD=(AC^2+AD^2-2AC*AD*cos100)^0.5
=(( 2sin50)^2+1-2*2sin50*1*cos100)^0.5
=(4sin50(sin50-cos100)+1)^0.5
=(4sin50(2sin30cos20)+1)^0.5
=(4cos40cos20+1)^0.5
=(2+2cos20)^0.5
=2cos10
设∠CDB=x,sinx/BC=sin140/CD
sinx=sin40(2sin50-1)/2cos10
=(sin100-sin40)/2cos10
=2sin30cos70/2cos10
=sin20/2cos10
=sin10
x<180-140=40
x=10