This link shows the derivation of this formula for a concave quadrilateral.
This link shows the derivation of this formula for a convex quadrilateral.
The formula I'm talking about is:
$$
\frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right]
$$
This post will be dedicated to the proving of this formula for any self intersecting quadrilateral.
Here, $ABCD$ is our self intersecting quadrilateral whose area will be denoted by $ \Delta $. We will be taking $AB=a, BC=b, CD=c, AD=d$ and $\angle B$ as $\theta$. We are assuming that $a,b,c,d,\theta$ are given. Also, we have to be careful while assigning these variables since the angle $\theta$ must lie between the sides $a,b$. $$ \Delta = Ar\triangle ADC+Ar\triangle ABC \\ \Delta =\frac{1}{2}cd\sin \angle D+\frac{1}{2}ab\sin \theta $$ Note that the intersecting area of $\triangle AEC$ is accounted for by $\sin \theta$ as it can be negative when $\theta$ is reflex. Now, by using the law of cosines, $$ AC^2=a^2+b^2-2ab\cos \theta \\ AC^2=c^2+d^2-2cd\cos \angle D $$ Subtracting these equations from each other yields: $$ a^2+b^2-2ab\cos \theta-c^2-d^2+2cd\cos \angle D=0 \\ \cos \angle D=\frac{c^2+d^2-a^2-b^2+2ab\cos \theta}{2cd} $$ Now, using the identity $ \sin\angle D= \sqrt{1-\cos^2\angle D} $, we get: $$ \sin\angle D= \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} $$ And putting this into $ \frac{1}{2}ab\sin\theta+\frac{1}{2}cd\sin\angle D $, we get: $$ \frac{1}{2}\left[ab \sin \theta + cd \sqrt{1- \left(\frac {c^2+d^2-a^2-b^2+2ab\cos\theta}{2cd}\right)^2} \right] $$ So, my formula is correct for any self intersecting quadrilateral. And, as seen in the linked posts, it is also correct for any concave and convex quadrilateral. This makes it a formula for a general quadrilateral which requires only four sides and one angle. Please let me know if I'm wrong.
The more important thing here is that the formula may be very tedious and more time consuming to use than the original method itself, but the fact that it exists shows that the area of a quadrilateral will always be fixed given the same four sides and one angle (as long as the given angle lies between two sides of the same length). I did not previously think that this could be the case.
Furthermore, this formula is more minimalistic than Bretschneider's formula because it requires one angle instead of two.
