Tolong jawab ya guys

Balok ABCD.EFGH

AB = 8 cm
BC = CG = 4cm

Perhatikan ∆AFH
AF = FH = √(AB² + BF²) = √(8² + 4²) = 4√5 cm

AH = diagonal sisi ADHE = 4√2 cm

O tengah AH
OA = 1/2 AH = 2√2 cm

FO = √(EO² + EF²) = √(8 + 8²) = 6√2 cm

Perhatikan ∆EFO siku di E
sudut antara EF dan AFH = sudut EFO = a

cos a = EF/FO
cos a = 8/(6√2)
cos a = 4/(3√2)
cos a = 4/(3√2) . √2 /√2
cos a = 4/6 √2
cos a = 2/3 √2