Then AE⊥CD of BE⊥CD.
∴∠AEB is the plane angle of dihedral angle A-CD-B.
The side length of regular tetrahedron A-BCD is 1.
∴, AE=√3/2 in regular triangle ACD.
In regular triangle BCD, BE=√3/2.
And AB= 1,
∴, cos∠ AEB = (AE 2+BE 2-AB 2)/(2AE * BE) =1/3 in △ABE.
∴ The plane cosine of dihedral angle A-CD-B is 1/3.
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