We can find angle BCA in the large triangle thusly :
sin BCA / 5.6 = sin 50 / 9.3
sin BCA = 5.6sin 50 / 9.3
So
arcsin (5.6 sin 50 / 9.3) = BCA = 27.47°
So....in the smaller triangle BDC, angle BDC is supplemental to angle BDA = 180 - 78 = 102°
So x = angle DBC =
180 - BDC - BCA =
180 - 102 - 27.47 =
50.53° = 50.5°