Explanation:
we have to imagine an inner right-angled triangle :
CE is the Hypotenuse (the side opposite to the 90° angle).
then the height of the pyramid (from the center of the base area to E) is one leg.
and the connection from C to the center of the base area is the second leg. that connection is half of the diagonal of the rectangular base area.
the angle of CE with the plane ABCD is simply the angle C of that right-angled triangle.
ok, we need a few things to be able to calculate that triangle.
let's start with the length of the (half) of the diagonal of the rectangular base area (we see that all sides of the base area are equal, so it is actually a square, but it does not really matter) .
Pythagoras helps us :
diagonal² = 14² + 14² = 196 + 196 = 392
diagonal = sqrt(392) = sqrt(49×4×2) = 7×2×sqrt(2) =
= 14×sqrt(2) cm
diagonal/2 = 14×sqrt(2)/2 = 7×sqrt(2) cm
and with Pythagoras we get now also CE itself :
CE² = 13² + (7×sqrt(2))² = 179 + 49×2 = 179 + 98 = 277
CE = sqrt(277) cm
since we have a right-angled triangle, we know the legs are sine and cosine of C (multiplied by CE).
so,
13 = sin(C) × sqrt(277)
sin(C) = 13 / sqrt(277) = 0.781094299...
C = 51.36087783...° ≈ 51.4°
the angle between line CE and the plane ABCD is
51.4°.