We can assume the kite is simetrycal along line WY, as such
[tex]ZW=XW=10[/tex][tex]m\angle\text{XSW}=90[/tex]
because the diagonals of a kite are perpendicular.
Since m∠XSW is 90°, triangle WSX is a right triangle. Thus
[tex]m\angle WXZ=m\angle WXS=180-90-46=44[/tex]
By the symmetry we discussed earlier,
[tex]m\angle WYZ=m\angle WYX=18[/tex]
And finally
[tex]m\angle XYZ=m\angle WYZ+m\angle WYX=18+18=36[/tex]
