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).

enter image description here

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 $780$:

enter image description here

$$0.5 \cdot 29.1 \cdot 27.75 \cdot \sin x+0.5 \cdot 27.8 \cdot 27.4 \cdot \sin y=780$$

$$29.1^2+27.75^2-2 \cdot 29.1 \cdot 27.75 \cdot \cos x=27.4^2+27.8^2-2 \cdot 27.4 \cdot 27.8 \cdot \cos y$$

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:

enter image description here

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.

enter image description here


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.


A small update with the two possible solutions:

enter image description here

  • 1
    $\begingroup$ You can rewrite the equations as $\sin x=b-a\sin y$, $\cos x=c+a \cos y$. Square both, add together and you'll get something like $A\sin y + B\cos y=C$ which is not that difficult to solve. $\endgroup$
    – Vasili
    Jan 18, 2023 at 14:38
  • 1
    $\begingroup$ By Heron's formula, you can find the area of a triangle if you know the lengths of its 3 sides. Let length of line AD be x. AD cuts the quadrilateral into 2 triangles. We know the sum of the area of these two triangles and the area for a single triangle is known up to a function of x. This implies x is the solution of a polynomial. Once you know x, the cosine rule tells you how to find the angles of a triangle given its side lengths. $\endgroup$ Jan 20, 2023 at 17:59
  • 1
    $\begingroup$ You might want to mention explicitly that the side-lengths are to be used in the order given. $\endgroup$
    – Blue
    Jan 22, 2023 at 6:10

4 Answers 4


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 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

$$a_2 (1 - t^2) - (a_1 - a_3 t)^2 = (a_3^2 - a_2)t^2 + 2(a_1a_3)t -a_1^2 = 0$$


I'm going to derive expressions of angles using complex numbers.

To simplify description, we will assume:

  1. the quadrilateral is convex.

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

  3. 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 $\epsilon = \pm 1$.


I think that your approach is nice and natural.

The system you've got can be solved in a simple way which I believe your high school students can understand easily.

The system you've got is the following : $$\frac 12CA\cdot CD\sin C+\frac 12BA\cdot BD\sin B=780$$ $$CA^2+CD^2-2\cdot CA\cdot CD\cos C=BA^2+BD^2-2\cdot BA\cdot BD\cos B$$

As pointed out in the comments, this system can be solved easily because it can be written as $$\begin{align}\sin C&=-\color{red}{\frac{BA\cdot BD}{CA\cdot CD}}\sin B+\frac{2\times 780}{CA\cdot CD} \\\\\cos C&=+\color{red}{\frac{BA\cdot BD}{CA\cdot CD}}\cos B+\frac{CA^2+CD^2-BA^2-BD^2}{2\cdot CA\cdot CD}\end{align}$$ (The reason of isolating $\sin C$ ($\cos C$) is to eliminate $C$ using $\sin^2C+\cos^2C=1$.)

Now, let us prove the following claim (a proof of the claim is written at the end of this answer).

Claim : The solutions of the following system $$\sin x=-\color{red}p\sin y+q\tag1$$ $$\cos x=+\color{red}p\cos y+r\tag2$$ $$0\lt x\lt \pi\qquad\text{and}\qquad 0\lt y\lt \pi\tag3$$ where $p,q,r$ are constants satisfying $p\not=0,q\geqslant 0$ and $(q,r)\not=(0,0)$ are $$(x,y)=\bigg(\arccos\bigg(pMN+p\sqrt{(1-M^2)(1-N^2)}+r\bigg),\arccos M-\arccos N\bigg),$$ $$\bigg(\arccos\bigg(pMN-p\sqrt{(1-M^2)(1-N^2)}+r\bigg),2\pi-\arccos M-\arccos N\bigg)$$ where $$M=\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}},\qquad N=\frac{r}{\sqrt{q^2+r^2}}$$

Using the claim, I got $$(C,B)\approx (1.4395019,1.4931531),(1.6423184,1.7082117221)$$ i.e. $$(C,B)\approx (82.4773842^\circ, 85.551376^\circ),(94.09791739^\circ, 97.873322^\circ)$$ which are the same as your solutions.

After getting $B$ and $C$, you can find the other angles by solving another similar system

$$\begin{align}\sin A&=-\color{red}{\frac{DB\cdot DC}{AB\cdot AC}}\sin D+\frac{2\times 780}{AB\cdot AC} \\\\\cos A&=+\color{red}{\frac{DB\cdot DC}{AB\cdot AC}}\cos D+\frac{AB^2+AC^2-DB^2-DC^2}{2\cdot AB\cdot AC}\end{align}$$

Proof of the claim :

Since $\sin^2x+\cos^2x=1$, we get, from $(1)(2)$, $$(-p\sin y+q)^2+(p\cos y+r)^2=1$$ i.e. $$p^2(\underbrace{\sin^2y+\cos^2y}_{=1})-2pq\sin y+2pr\cos y+q^2+r^2=1$$ i.e. $$\frac{r}{\sqrt{q^2+r^2}}\cos y-\frac{q}{\sqrt{q^2+r^2}}\sin y=\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}}$$ which can be written as $$\cos(y+\alpha)=\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}}$$ where $$\alpha=\arccos\bigg(\frac{r}{\sqrt{q^2+r^2}}\bigg),\quad \cos\alpha=\frac{r}{\sqrt{q^2+r^2}},\quad \sin\alpha=\frac{q}{\sqrt{q^2+r^2}}\geqslant 0$$ Since $0\lt y\lt \pi$ and $0\leqslant \alpha\leqslant \pi$, we have $0\lt y+\alpha\lt 2\pi$, so $$y+\alpha=\arccos\bigg(\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}}\bigg),\quad 2\pi-\arccos\bigg(\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}}\bigg)$$

Hence, we get $$y_1=\arccos M-\arccos N, \qquad y_2=2\pi-\arccos M-\arccos N$$ where $$M=\frac{1-p^2-q^2-r^2}{2p\sqrt{q^2+r^2}},\quad N=\frac{r}{\sqrt{q^2+r^2}}$$

Finally, from $(2)$, we have $$x=\arccos(p\cos y+r)$$ and finally get $$\begin{align}x_1&=\arccos(p\cos(\arccos M-\arccos N)+r) \\\\&=\arccos\bigg(pMN+p\sqrt{(1-M^2)(1-N^2)}+r\bigg)\end{align}$$ and $$\begin{align}x_2&=\arccos(p\cos(2\pi-\arccos M-\arccos N)+r) \\\\&=\arccos(p\cos(\arccos M+\arccos N)+r) \\\\&=\arccos\bigg(pMN-p\sqrt{(1-M^2)(1-N^2)}+r\bigg)\quad\blacksquare\end{align}$$


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