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:
Best Answer
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.