The angle you want - call it θ - is such that
tan(θ) = FC / AC
Find the length of the diagonal AC, i.e. a diagonal of the rectangle ABCD. ABC forms a right triangle with legs AB = 70 and BC = 50, so
AC² = AB² + BC²
→ AC = √(70² + 50²) = 10 √74
Find FC using the given angle of the sloping face:
tan(30º) = FC / BC
→ 1/√3 = FC / 50
→ FC = 50/√3
Now solve for θ :
tan(θ) = (50/√3) / (10 √74)
→ tan(θ) = 5/√222
→ θ ≈ 18.6º
