The goal is to find the angle between the height of the pyramid and its front face. For this, we need to label the midpoint of YZ as A, and to find XA and UX, the segments opposite and adjacent ∠AUX. The answer will be ∠AUX= Arctan(XA/UX).
Since the base WVYZ is rectangular,△VYZ is a right triangle with hypotenuse VZ and legs VY=5 and YZ=6. So VZ^2 = VY^2 + YZ^2 = 25+36 = 61,
VZ = √61.
Since X is the midpoint of VZ,
VX = XZ = √61 / 2.
Since UX is perpendicular to the base, △UXV is a right triangle with hypotenuse UV=12, UV^2=144, and one known leg VX = √61 / 2, VX^2=61/4. The other leg, UX, is the height of the pyramid.
UX = √(UV^2 - VX^2) = √(144 - 61/4) = √515 / 2.
Label the midpoint of YZ as A. Clearly
XA ⟂ YZ, XA || VY, so
XA = VY/2 = 5/2.
△UXA is of course a right triangle, and the required angle between pyramid height UX and front face UZY is ∠AUX, located at the apex U. The side adjacent to ∠AUX is UX, and the opposite side is XA. The tangent of ∠AUX is opposite / adjacent,
tan(∠AUX) = XA/UX = (5/2) / (√515/2) = 5/√515
∠AUX = Arctan(tan(∠AUX)) = Arctan(5/√515) = 12.43°