Solution: Let A be the symmetry point A ′, A″ is about BC and CD, connecting A ″″, passing BC to M, passing CD to N, then A ″″ is the minimum circumference of △AMN. Let a be the extension of DA.

∫∠DAB = 120，

∴∠haa′=60，

∴∠aa′m+∠a″=∠haa′=60，

∠∠MA′A =∠MAA′，∠NAD=∠A″,

And ∠ Maa +∠MAA' =∠ Amn, ∠ Nader+∠ A "= ∠ ANM,

∴∠amn+∠anm=∠ma′a+∠maa′+∠nad+∠a″=2(∠aa′m+∠a″)=2×60 = 120，

So the answer is: 120.

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