Maths problem [Trigonometry]

(14) The diagonal lines AF and BE are equal = L
By Pythagoras theorem,
L^2 = 2^2 + 4^2 + 8^2 = 84
L = 2sqrt(21)

Distance of A to mid-point G of the cuboid = L/2 = sqrt(21)
tan@/2=half of AB/AG = 4/sqrt(21) (B)

(15) Let AB=X
AB/BC=tan30
BC = x/tan(30) = sqrt(3).x

BG/BC=tan60
BG=BC * sqrt(3)
= sqrt(3).x.sqrt(3)
=3x

tan@=BG/AB = 3x/x = 3 (D)

(20) BCE is an isoceles triangle since BC=BE
angle BCE= angle BEC=@, and
2@ + 60 = 180
@=60

So BCE is indeed equilateral => CE= 1m

AC=AE
AC^2=2^2+1^2=5
AC=sqrt(5)

by cosine rule
CE^2 = AC^2+AE^2-2(AC)(AE)cos@
1 = 5 + 5 -10cos@
10cos@=9
cos@=9/10 (D)