Can I make more assumptions?

Drop a perpendicular from $A$ to $BC$ at point $P.$ Next, drop a perpendicular from $D$ to $BC$ at point $Q.$ Finally, draw a line parallel to $BC$ through $D,$ and let $R$ be the point of intersection of this line and $AP.$

Then the area of the quadrilateral $ABCD$ is the sum of the areas of triangles $\Delta ABP, \Delta DQC, \Delta ARD$ and the rectangle $RPQD.$ Looking at these areas separately, we have:

(i) Area $\Delta ABP = \dfrac{1}{2}*(4\sin(75^{\circ}))(4\cos(75^{\circ})) = 4\sin(150^{\circ}) = 2$

(ii) Area $\Delta DQC = \dfrac{1}{2}(3\sin(75^{\circ}))(3\cos(75^{\circ})) = \dfrac{9}{4}\sin(150^{\circ}) = \dfrac{9}{8}$

(iii) Area $\Delta ARD = \dfrac{1}{2}(AR)(RD) = \dfrac{1}{2}(AR)(PQ) =$

$\dfrac{1}{2}((4 - 3)\sin(75^{\circ}))(8 - (4 + 3)\cos(75^{\circ})) =$

$\dfrac{1}{2}(8\sin(75^{\circ}) - 7\sin(75^{\circ})\cos(75^{\circ})) =$

$4\sin(75^{\circ}) - \dfrac{7}{4}\sin(150^{\circ}) = 4\sin(75^{\circ}) - \dfrac{7}{8}$

(iv) Area rectangle $RPQD = (RP)(PQ) = 3\sin(75^{\circ})(8 - 7\cos(75^{\circ})) =$

$24\sin(75^{\circ}) - \dfrac{21}{2}\sin(150^{\circ}) = 24\sin(75^{\circ}) - \dfrac{21}{4}.$

Adding these areas together gives us the area of $ABCD$ to be

$2 + \dfrac{9}{8} - \dfrac{7}{8} - \dfrac{21}{4} + 28\sin(75^{\circ}) = 28\sin(75^{\circ}) - 3.$

Now $\sin(75^{\circ}) = \sin(45^{\circ} + 30^{\circ}) = \dfrac{\sqrt{2}}{2}\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\right) = \dfrac{\sqrt{2} + \sqrt{6}}{4}.$

Thus the final area calculation is $\boxed{7\sqrt{6} + 7\sqrt{2} - 3}.$

two Triangles and a trapezoid :) now you have it!

Extend lines AB and DC to intersect at point E. Find area of triangle EBC. Find area of triangle EAD. Subtract area of triangles EAD from EBC.

TADAAAA

Is harder than it sounds to get an exact value... but is fun