Call the intersection point of the two diagonals X.
Looking at Triangle BCX: ∠BCX = 85°, BX = 10 and CX = 7.
Law of Sines for Triangle BCX: sin(∠BCX) / BX = sin(∠CBX) / CX
---> sin(85°) / 10 = sin(∠CBX) / 7
---> 7 x sin(85°) / 10 = sin(∠CBX)
1) Find ∠CBX.
2) Use ∠CBX and ∠BCX to find ∠BXC.
3) Use ∠BXC to find ∠BXA.
4) Use the Law of Cosines on Triangle AXB to find AB.
5) Use the Law of Sines on Triangle AXB to find ∠ABX.
6) Find ∠ABC.
7) Find ∠BAD.