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Answers to mathematical classics
Solution: Let the side length of a square be 4K.

Square ABCD

∴AB=BC=CD=AD=4K,∠B=∠C=∠D=90

∫BE = BC/2

∴BE=2K

∴CE=BC-BE=2K

∫CF = CD/4

∴CF=K

∴DF=CD-CF=3K

∴AE? =AB? +BE? = 16K? +4K? =20K?

EF? =CE? +CF? =4K? +K? =5K?

AF? =AD? +DF? = 16K? +9K? =25K?

∴AE? +EF? =AF? =25K?

∴∠AEF=90

∴sin∠EAF=EF/AF=√(5K? )/√(25K? )=√5/5

cos∠EAF=AE/AF=√(20K? )/√(25K? )=2√5/5

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