trigonometry question

(a)

let F' be the projection of F on BC
so the required angle is ∠FAF'

in triangle BFF', sin 45 = FF'/8 => FF' = 8sin 45
in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113
in triangle AFF', sin∠FAF' = FF'/AF
=> sin∠FAF' = 8sin 45/sqrt113 => ∠FAF' = 32.2

Therefore, the angle between FA and plane ABCD is 32.2

(b)

in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113
in triangle ABC, AC^2 = 7^2 + 8^2 => AC = sqrt113
in triangle FBC, FC^2 = 8^2 + 8^2 - 2(8)(8)cos 45=> FC = 6.12
in triangle AFC, cos∠FAC = (AF^2 + AC^2 - FC^2)/[2(AF)(AC)]
=>∠FAC = 33.5

Therefore, the angle between FA and AC is 33.5