4. Pentagon $A B C D E$ is inscribed in a circle of radius $R$. It is known that $\angle B=110^{\circ}, \angle E=100^{\circ}$. Find the side $C D$.
Solution. The degree measures of arcs $\overline{A D}$ and $\overline{C A}$ are respectively $2 \cdot$ $\left(180^{\circ}-\angle E\right)$ and $2 \cdot\left(180^{\circ}-\angle B\right)$. The sum of the degree measures of arcs $\overline{A D}, \widetilde{C A}$, and $\overline{D C}$ is $360^{\circ}$. Therefore, the measure of angle $C A D$ (which is half the degree measure of arc $\overline{D C}$) is determined by the equation ${ }^{1} \quad \angle C A D=\angle B+\angle E-180^{\circ}=30^{\circ}$. Using the formula for the radius of the circumcircle of triangle $C A D$, we find ${ }^{2}$

![](https://cdn.mathpix.com/cropped/2024_05_06_cd78e7c3cd59bc2a216dg-2.jpg?height=323&width=302&top_left_y=1140&top_left_x=1616)

$$
R=\frac{C D}{2 \cdot \sin \angle C A D}=C D
$$

Answer: $C D=R$.

## Comments.

- ${ }^{1}$ From this, it follows that the sum of any two non-adjacent angles of an inscribed pentagon is greater than $180^{\circ}$.
- ${ }^{2}$ There are analogous formulas:

$$
-2 R=\frac{C D}{\sin (\angle B+\angle E)}=\cdots=\frac{B C}{\sin (\angle A+\angle D)}
$$

It turns out that these equalities express the necessary and sufficient condition for a circumcircle of radius $R$ to be described around a given pentagon.