Answer:
∠HAF ≈ 64.90°
Explanation:
You want the angle HAF, a base angle in the isosceles triangle formed by the face diagonals of the cuboid with dimensions 6 cm, 6 cm, and 8 cm.
Diagonal lengths
The lengths of the diagonals can be found using the Pythagorean theorem:
HA² = HD² + DA²
HA² = 6² +6² = 2·6²
HA = 6√2 . . . cm
FA² = FB² +BA²
FA² = 6² +8² = 36 +64 = 100
FA = 10 . . . cm
Diagonal FH is the diagonal of a rectangle with the same dimensions as the rectangle whose diagonal is FA:
FH = FA = 10 . . . cm
Law of Cosines
The angle A can be found using the law of cosines:
FH² = FA² +HA² -2·FA·HA·cos(A)
A = arccos((FA² +HA² -FH²)/(2FA·HA))
A = arccos((100 +72 -100)/(2·10·6√2)) = arccos(72/(120√2))
A = arccos(0.3√2) ≈ 64.90°
Angle HAF is about 64.90°.