What percentage error do we make if we take half of the chord corresponding to the approximate $120^{\circ}$ central angle of a regular heptagon?
The central angle corresponding to the side of a regular heptagon is $\frac{360^{\circ}}{7}$, and half of this is $\alpha=\frac{360^{\circ}}{14}=25^{\circ} 42^{\prime} 51^{\prime \prime}$ (see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_bdbf466dba86568a1890g-1.jpg?height=528&width=544&top_left_y=259&top_left_x=777)

If we choose the radius of the circle as the unit and denote the side by $a$, then in the right-angled triangle $AFO$,

$$
\frac{a}{2}=\sin \alpha
$$

from which

$$
a=2 \sin \alpha
$$

Similarly, we calculate half of the chord corresponding to a $120^{\circ}$ central angle in the same radius circle:

$$
a^{\prime}=\sin 60^{\circ}
$$

The percentage error is obtained by dividing the absolute value of the deviation from the correct value by the hundredth part of the correct value.

$$
\begin{gathered}
h=\frac{\left|a-a^{\prime}\right|}{\frac{a}{100}}=100\left|1-\frac{a^{\prime}}{a}\right|=100\left|1-\frac{\sin 60^{\circ}}{2 \sin 25^{\circ} 42^{\prime} 51^{\prime \prime}}\right| \sim \\
\sim 100\left|1-\frac{0.8660}{0.8678}\right| \sim 100|1-0.998|=0.2 \%
\end{gathered}
$$

We also noticed that the correct value is larger than the approximate value.