<s>[INST] Compute the area of Quadrilateral $ABCD$ As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:

I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution (which I will post in a comment below since I don't want to clutter the question) is a little messy and not efficient. Are there any better ways to do this? Geometric and/or trigonometric approaches are all welcomed!
EDIT: I have posted my solution below!
 [/INST] 
Let the intersection of the two diagonals be $O$.
And let $ AO = x , OC = y, AD = z $
Then it follows from the law of sines that
$ DO = a x \hspace{25pt}$ where $ a = \dfrac{\sin(105^\circ)}{\sin(30^\circ) } $
and
$ OB = b y \hspace{25pt}$ where $ b = \dfrac{\sin(75^\circ)}{\sin(60^\circ)} $
Apply the law of cosines to $\triangle AOD$, $\triangle OBD$, $\triangle ABC $ gives us the following three quadratic equations:
$ z^2 = x^2 (1 + a^2 - 2 a \cos(45^\circ) )$
$ z^2 = y^2 (1 + b^2 - 2 b (\cos(45^\circ) )$
$ z^2 + (x + y)^2 - 2 z (x + y) \cos(75^\circ) = 2^2 = 4 $
Using a quadratic system solver (for example from Wolframalpha.com), we get:
$ x = \sqrt{3} - 1 $.
$ y = 3 - \sqrt{3} $.
$ z = \sqrt{2} x = \sqrt{6} - \sqrt{2} $
Now the area is given by
$ \text{Area} = \dfrac{1}{2} \sin(45^\circ) \left( a x^2 + b y^2 + xy (a + b) \right) $
Evaluating the above expression, yields
$\text{Area} = 2$
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