设∠D=n,∠ACD=80-n
在△ABC里,AC/sin∠ABC=BC/sinA
在△ADC里,AC/sinD=AD/sin∠ACD
由于AD=BC,上下两式相除,sinD/sin∠ABC=sin∠ACD/sinA
得到:sinn/sin40=sin(80-n)/sin100
sinn/sin40=(sin80cosn-cos80sinn)/sin80
1/sin40=ctgn-ctg80
ctgn=ctg80+1/sin40
=cos80/sin80+1/sin40
=(2cos40^2-1)/2sin4ocos40+1/sin40 cos40^2=(cos40)^2,下同
=(1+cos40)/sin40-1/sin80
=2cos20^2/2sin20cos20-1/sin80
=cos20/sin20-1/cos10
=(cos20cos10-sin20)/sin20cos10
=(cos20cos10-2sin10cos10)/sin20cos10
=(cos20-2sin10)/sin20
=(cos20-cos80-sin10)/sin20
=(2sin50sin30-sin10)/sin20
=(sin50-sin10)/sin20
=2cos30sin20/sin20
=2cos30=√3
n=30°
三角函数参考地址:http://www.wen8.net/science/maths/3jiaohs.htm