Finding all four angles in a quadrilateral given four sides and its area I am creating a small investigation for my students and I'd like to check whether my mark scheme is correct.
In one of the questions, students are asked to find the angles of a quadrilateral with sides $27.4$, $27.8$, $27.75$ and $29.1$ knowing also that the area is $780$ (image below).

Possible solution
By splitting the quadrilateral into two triangles using the diagonal $AD$, we can build two equations: one that, through the cosine rule states that the diagonal is equal whether you use triangle $1$ or $2$, and the other that takes into account that the area is $780$:

$$0.5 \cdot 29.1 \cdot 27.75 \cdot \sin x+0.5 \cdot 27.8 \cdot 27.4 \cdot \sin y=780$$
$$29.1^2+27.75^2-2 \cdot 29.1 \cdot 27.75 \cdot \cos x=27.4^2+27.8^2-2 \cdot 27.4 \cdot 27.8 \cdot \cos y$$
The thing about this method is that it is very hard to find the solution to this linear system. The way that I did that was to input it on Wolfram, which gave me the following result:

This means that angle x is approximately $94.0979^\circ$ (the other angles can be found using a similar approach).
The link to this animation in Geogebra can be found here.

Question:
Is there a simpler way that this can be done (the main problem is solving that system)? Assume that the students will have a calculator and access to Geogebra.
Update
A small update with the two possible solutions:

 A: This is not a complete solution but I wanted to propose a different approach. We can find one of the diagonals using Heron's formula. Once diagonal is known, the angles can be determined using law of cosines.
$P_1=27.4+27.8+x=55.2+x; P_2=27.75+29.1+x=56.85+x$ $$4 \cdot 780=\sqrt{(55.2+x)(x+0.4)(x-0.4)(55.2-x)}+\sqrt{(56.85+x)(x+1.35)(x-1.35)(56.85-x)}=\sqrt{(55.2^2-x^2)(x^2-0.16)}+\sqrt{(56.85^2-x^2)(x^2-1.35^2)}=\sqrt{3074.2x^2-x^4-487.5264}+\sqrt{3233.745x^2-x^4-5890.17875625}$$
This leads to a quadratic equation for $x$. Still a bit ugly but doable with a calculator.
A: A solution that only requires solving quadratic equations and some trigonometric identities.
Using the Bretchneider's formula for the area of the quadrilateral
$$
A = \sqrt{(s - a)(s - b)(s - c)(s - d) - abcd \cos\left(\theta \right)^2}
$$
where $a,b,c,d$ are the sides and $s=(a+b+c+d)/2$
This will give two solutions $\theta_1$ and $\theta_2$, then using $x = \theta + \delta$, $y=\theta - \delta$ and the equation for the diagonal you get
$$ a^2+b^2 - 2ab \cos(\theta - \delta) = c^2 + d^2 - 2cd \cos(\theta + \delta) $$
Expanding the terms $\cos(\theta \pm \delta)$ and using $cos(\delta) = \sqrt{1 - \sin(\delta)^2}$
$$ a^2+b^2 - c^2 - d^2 - 2(ab - cd) \cos(\theta)\sqrt{1 - \sin(\delta)^2} - 2(ab + cd)\sin(\theta)\sin(\delta) = 0$$
Replacing constants $a_1, a_2, a_3$, and variable $t=\sin(\delta)$, we have
$$ a_1 - a_2\sqrt{1 - t^2} - a_3t = 0$$
Solutions can be found by solving the quadratic equation
$$a_2 (1 - t^2) - (a_1 - a_3 t)^2 = (a_3^2 - a_2)t^2 + 2(a_1a_3)t -a_1^2 = 0$$
A: I'm going to derive expressions of angles using complex numbers.
To simplify description, we will assume:

*

*the quadrilateral is convex.


*the vertices $A,B,C,D$ of quadrilateral are placed counterclockwisely along its perimeter.


*instead of $\alpha,\beta,\gamma,\delta$, we will use $\angle A, \angle B, \angle C,\angle D$ to denote the interior angle of quadrilateral at corresponding vertex.
The labeling of vertices in question is incompatible with this assumption.
I will relabel the vertices $C, D, B$ there to $B, C, D$. Under this relabelling, the angles $\angle A, \cdots, \angle D$ here corresponds to the old usage of $\alpha,\cdots,\delta$ in question body.
To proceed further, identity the plane with complex plane $\mathbb{C}$, define

*

*complex numbers $\alpha = B-A, \beta = C-B, \gamma = D-C, \delta = A- D$.

*$a = |\alpha|, b = |\beta|, c = |\gamma|, d = |\delta|$ and $\ell = |\alpha+\beta|$.

*$\Delta_1 = 2\verb/Area/(ABC)$, $\Delta_2 = 2\verb/Area/(CDA)$ and
$\Delta = \Delta_1 + \Delta_2 = 2\verb/Area/(ABCD)$

*$P_{ab} = -P_{cd} = \frac12(a^2+b^2 - c^2-d^2)$ and $P_{bc} = -P_{da} = \frac12(b^2+c^2 - d^2-a^2)$.

*$Q_{ab} = Q_{cd} = \sqrt{\Delta^2 + P_{ab}^2}$ and $Q_{bc} = Q_{da} = \sqrt{\Delta^2 + P_{bc}^2}$
If one apply sine and cosine rules to triangles $ABC$ and $CDA$, it is not hard
to see
$$\begin{align}
\beta\bar{\alpha}  &= \frac12(\ell^2 - a^2 - b^2) + \Delta_1 i\\
\delta\bar{\gamma} &= \frac12(\ell^2 - c^2 - d^2) + \Delta_2 i\\
\end{align}$$
Taking complex conjugate of $2^{nd}$ equation and subtract it from $1^{st}$ equation, we get
$$\beta\bar{\alpha} - \gamma\bar{\delta} = i\eta\quad\text{ where }\quad
\eta \stackrel{def}{=} \Delta + P_{ab}i$$
Since the quadrilateral is convex and we label its vertices counterclockwisely,
$\Im (\beta\bar{\alpha}) > 0$ and $\arg(\beta\bar{\alpha})$ is the exterior angle at $B$. Let
$$\theta = \arg\left(\frac{i\eta}{\beta\bar{\alpha}}\right) = \arg\left(1 - \frac{\gamma\bar{\delta}}{\beta\bar{\alpha}}\right),$$
we have
$$\angle B = \pi - \arg(\beta\bar{\alpha})
= \pi - \arg{i\eta} + \arg\left(\frac{i\eta}{\beta\bar{\alpha}}\right)
= \frac{\pi}{2} - \arg{\eta} + \theta
$$
Consider the triangle in $\mathbb{C}$ formed by $0$, $\beta\bar{\alpha}$ and $i\eta$. $|\theta|$ is simply the angle of this triangle at $0$.
We can compute it using cosine rules:
$$|\theta| = \cos^{-1}\left(\frac{|\eta|^2 + a^2b^2 - c^2d^2}{2ab|\eta|}\right)$$
$\theta$ is positive/negative depends on whether $0,\beta\bar{\alpha}, i\eta$ is ordered counterclockwisely/clockwisely along the triangle's perimeter.
So there are two possible solutions $\angle B$ and they having the form:
$$\angle B = \frac{\pi}{2} - \tan^{-1}\frac{P_{ab}}{\Delta} + \epsilon_B \cos^{-1}\left(\frac{Q_{ab}^2 + a^2b^2 - c^2d^2}{2abQ_{ab}}\right)$$
and $\epsilon_B = {\rm sign\;}\theta
= {\rm sign\;}\Im\left(1 - \frac{\gamma\bar{\delta}}{\beta\bar{\alpha}}\right)
$ can take values $\pm 1$.
The other angles can be obtained by cyclic permutation of vertices. It is not hard to verify $\epsilon_A = -\epsilon_B = \epsilon_C = -\epsilon_D$. The two set of solutions for the angle are
$$\begin{align}
\alpha_{\rm old} = \angle A &= \frac{\pi}{2} 
+ \tan^{-1}\frac{P_{bc}}{\Delta} 
- \epsilon \cos^{-1}\left(\frac{Q_{bc}^2 + d^2a^2 - b^2c^2}{2daQ_{bc}}\right)\\
\beta_{\rm old} = \angle B &= \frac{\pi}{2} 
- \tan^{-1}\frac{P_{ab}}{\Delta} 
+ \epsilon \cos^{-1}\left(\frac{Q_{ab}^2 + a^2b^2 - c^2d^2}{2abQ_{ab}}\right)\\
\gamma_{\rm old} = \angle C &= \frac{\pi}{2}
- \tan^{-1}\frac{P_{bc}}{\Delta}
- \epsilon \cos^{-1}\left(\frac{Q_{bc}^2 + b^2c^2 - d^2a^2}{2bcQ_{bc}}\right)\\
\delta_{\rm old} = \angle D &= \frac{\pi}{2}
+ \tan^{-1}\frac{P_{ab}}{\Delta}
+ \epsilon \cos^{-1}\left(\frac{Q_{ab}^2 + c^2d^2 - a^2b^2}{2cdQ_{ab}}\right)
\end{align}
$$
for $\epsilon = \pm 1$.
