求x多少度
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用正弦余弦定理，然后进行三角函数的化简。
设AB=1，BD=CD=sin80°/sin40°=2cos40°
AD=sin60°/sin40°，AE=sin80°/sin30°=2sin80°
DE²=AD²+AE²-2AD.AE.cos10°
=sin²60°/sin²40°+4sin²80°-2.sin60°/sin40°.2sin80°cos10°
=3/4sin²40°+4sin²80°（1-sin60°/sin40°）
=3/4sin²40°-4sin²80°（sin60°-sin40°）/sin40°
=3/4sin²40°-4sin²80°2cos50°sin10°/sin40°
=3/4sin²40°-8sin²80°sin10°
=3/4sin²40°-4sin80°sin160°
=3/4sin²40°-4sin80°sin20°
=3/4sin²40°-2（cos60°-cos100°）
=3/4sin²40°-1-2cos80°）
=（3-4sin²40°-8sin²40°cos80°）/4sin²40°
=（3-2（1-cos80°）-4（1-cos80°）cos80°）/4sin²40°
=（1+2cos80°-4cos80°+4cos²80°）/4sin²40°
=（1-2cos80°+4cos²80°）/4sin²40°
=（1-2cos80°+2（1+cos160°））/4sin²40°
=（3-2cos80°-2cos20°）/4sin²40°
=（3-2（cos80°+cos20°））/4sin²40°
=（3-4cos50°cos30°）/4sin²40°
=（3-2√3cos50°）/4sin²40°
=√3（√3/2-cos50°）/2sin²40°
=√3（cos30°-cos50°）/2sin²40°
=2√3sin40°sin10°/2sin²40°
=√3sin10°/sin40°
=2sin60°sin10°/sin40°
=（cos50°-cos70°）/sin40°
=1-sin20°/2sin20°cos20°
=（1-1/2cos20°）
=（cos20°-cos60°）/cos20°
=2sin40°sin20°/cos20°
=4sin20°sin20°
=4sin²20°
DE=2sin20°
cosx=（DE²+AE²-AD²）/2DE.AE
=（4sin²20°+4sin²80°-sin²60°/sin²40°）/[2.2sin20°.2sin80°]
=（4sin²20°+4sin²80°-3/4sin²40°）/[8sin20°sin80°]
=[2(1-cos40°]+2（1-cos160°）-3/4sin²40°]/[8sin20°sin80°]
=[4-2cos40°-2cos160°-3/4sin²40°]/[8sin20°sin80°]
=[4-2cos40°+2cos20°-3/4sin²40°]/[8sin20°sin80°]
=[4+2(cos20°-cos40°)-3/4sin²40°]/[8sin20°sin80°]
=[4+4sin30°sin10°-3/4sin²40°]/[8sin20°sin80°]
=[4+2sin10°-3/4sin²40°]/[8sin20°sin80°]
=[16sin²40°+8sin10°sin²40°-3]/[32sin20°sin80°sin²40°]
=[8(1-cos80°）+4sin10°（1-cos80°）-3]/[32sin20°sin80°sin²40°]
=[5-8cos80°+4cos80°（1-cos80°）]/[32sin20°sin80°sin²40°]
=[5-8cos80°+4cos80°-4cos²80°）]/[32sin20°sin80°sin²40°]
=[5-4cos80°-2（1+cos160°）]/[32sin20°sin80°sin²40°]
=[3/4-cos80°+1/2.cos20°]/[8sin20°sin80°sin²40°]
=[cos²30°-cos80°+sin30°cos20°）/[8sin20°sin80°sin²40°]
=[cos²30°-cos80°+（1/2）（sin50°+sin10°）】/[8sin20°sin80°sin²40°]
=[cos²30°-1/2.sin10°+（1/2）sin50°】/[8sin20°sin80°sin²40°]
=[cos²30°+（1/2）（sin50°-sin10°）】/[8sin20°sin80°sin²40°]
=[cos²30°+cos30°sin20°】/[8sin20°sin80°sin²40°]
=cos30°（cos30°+sin20°）/[8sin20°sin80°sin²40°]
=cos30°（sin60°+sin20°）/[8sin20°sin80°sin²40°]
=2cos30°sin40°cos20°/[8sin20°sin80°sin²40°]
=cos30°cos20°/[4sin20°sin80°sin40°]
=cos30°cos20°/[2（cos20°-cos60°）sin80°]
=cos30°cos20°/[2cos20°sin80°-sin80°]
=cos30°cos20°/[sin100°+sin60°-sin80°]
=cos20°
x=20°
