Quadrilateral area from three sides and two angles I know there is already a question about resolving a quadrilateral from three sides and two angles, but I want to ask about a special case.  Firstly, two of the sides are known to be of equal size.  Secondly, I'm only interested in the area, not in the remaining angles or lengths.  Can anyone suggest a simple formula?

 A: Here's a brute-force derivation ...

$$\begin{align}
|\square PQRS| &= |\triangle PQR| + |\triangle PRS| \\[4pt]
&=\frac12 p q \sin \angle Q + \frac12 d r \sin(\angle R-\angle PRQ) \\[4pt]
&=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot d \cos\angle PRQ - \cos\angle R\cdot d\sin\angle PRQ\;\right) \;\right)\\[4pt]
&=\frac12 \left(\;p q \sin \angle Q + r \left(\;\sin\angle R\cdot(q-p\cos\angle Q) - \cos\angle R\cdot p\sin\angle Q\;\right) \;\right)\\[4pt]
&=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \left(\;\sin\angle R \cos\angle Q + \cos\angle R\sin\angle Q\;\right) \;\right)\\[16pt]
&=\frac12 \left(\;p q \sin \angle Q +qr\sin\angle R - pr \sin(\angle Q+\angle R)\;\right)\\
\end{align}$$

In the specific case of $p=r$, the formula can be manipulated thusly:
$$\begin{align}
|\square PQRS| &= \frac{p}{2}\;\left(\;q(\sin Q + \sin R) - p \sin(Q+R) \;\right) \\[4pt]
&= p\sin\frac{Q+R}{2}\;\left(\;q\cos\frac{Q-R}{2}- p \cos\frac{Q+R}{2} \;\right)
\end{align}$$
A: We can try using coordinate geometry and determinants to find the area.

WLOG, suppose $u=AB$ lie on the $x$-axis with $A=(0,0)$ and $B=(u,0)$. We can drop a line perpendicular to the $x$-axis through $E$, and we have a right triangle. Since $L=EA$, we can easily write points $E,F$ in $xy$-coordinate space. And thus we have the following vertices:
$$A=(0,0)\\B=(u,0)\\E=(L \cos a,L\sin a)\\
F=(u-L\cos b,L\sin b)$$

Let's split the quadrilateral into two triangles: $\triangle AEF$ and $\triangle ABF$. From here, we can write the area as:
$$A_{ABEF}=\frac12\left|
\begin{array}{ccc}
 0 & 0 & 1 \\
 u & 0 & 1 \\
 u-L \cos (b) & L \sin (b) & 1 \\
\end{array}
\right|-\frac{1}{2}\left| 
\begin{array}{ccc}
 0 & 0 & 1 \\
 L \cos (a) & L \sin (a) & 1 \\
 u-L \cos (b) & L \sin (b) & 1 \\
\end{array}
\right|$$
Which simplies into:
$$A_{ABEF}=\frac{1}{2} L \,u (\sin (a)+\sin (b))-\frac{1}{2} L^2 \sin (a+b)$$

Note: it is important to consider the signs of the resulting $\cos$ of angles $a,b$ and of the determinants in the derivation of the area.
