A quadrilateral has two adjacent sides of lengths 2 and 1 units, respectively, and the included angle between them is $60^{\circ}$. The quadrilateral is both a cyclic and a tangential quadrilateral. What are the lengths of the other two sides of the quadrilateral?
Solution. In the quadrilateral $A B C D$, $A B=2, B C=1$ and $\angle A B C=60^{\circ}$. Let the side $A D$ be $x$; let $C D=y$ and $A C=d$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bb678460c86b4ef5f1ddg-1.jpg?height=501&width=528&top_left_y=259&top_left_x=793)

Since the quadrilateral is a tangential quadrilateral,

$$
1+x=2+y
$$

from which $y=x-1$. Since $A B C D$ is also a cyclic quadrilateral, $\angle A D C=120^{\circ}$. Let's apply the cosine rule to triangle $A C D$:

$$
d^{2}=x^{2}+y^{2}-2 x y \cos 120^{\circ}
$$

Substituting the values of $y$ and $\cos 120^{\circ}$:

$$
d^{2}=x^{2}+(x-1)^{2}-2 x(x-1) \cdot\left(-\frac{1}{2}\right)
$$

The value of $d^{2}$ can be calculated from triangle $A B C$ using the cosine rule: $d^{2}=1^{2}+2^{2}-2 \cdot 1 \cdot 2 \cdot \frac{1}{2}$, from which $d^{2}=3$. Substituting this into (1) results in the following quadratic equation:

$$
3 x^{2}-3 x-2=0
$$

from which $x=\frac{\sqrt{33}+3}{6} \approx 1.457$, and the other side of the quadrilateral is $y \approx 0.457$ units.