In a unit circle, we draw the diameter $A C$, and let $B$ be the point on the arc $A C$ such that $\angle B O A = 30^{\circ}$ (where $O$ is the center of the circle). The tangent at $A$ intersects the line $C B$ at $D$. What is the difference between the length of the arc $A B$ and the segment $A D$?
Since the arc $\widehat{A B}$ is $30^{\circ}$, the corresponding inscribed angle is $15^{\circ}$. Thus, from the right-angled triangle $A D C$,

$$
A D=A C \operatorname{tg} 15^{\circ}=2 \operatorname{tg} 15^{\circ}
$$

The length of the arc $\widehat{A B}$ is $\widehat{\alpha}$, where $\widehat{\alpha}$ is the value of the angle $\alpha=30^{\circ}$ expressed in radians, i.e., $\frac{\pi}{6}$. Thus, the difference is $\left|\frac{\pi}{6}-2 \cdot \operatorname{tg} 15^{\circ}\right|$. Performing the calculation to four decimal places, the result is $\left|\frac{3.1415}{6}-2 \cdot 0.2679\right|=0.0123$. The difference is smaller than one fortieth of the arc.

The accuracy of the result naturally depends on the number of decimal places we calculated to. This is how it happened that some submissions, which were calculated to only one decimal place, concluded that there was no difference between the two distances. These submissions did not receive points.

It was correct to note that - since the expression contains $\pi$ and $\sqrt{3}$ - we cannot obtain an exact value, only an approximate one. The question now is just how accurately we need to calculate. Generally, practice provides the answer. For example, if it is about manufacturing a precision instrument, then even hundredths or thousandths of a millimeter can play a role. But if I want to calculate how much wallpaper is needed for a room, I will calculate to the nearest $1 \mathrm{dm}$, since wallpaper is usually bought by the meter and not by the centimeter. We can read a lot of interesting things about approximate calculations and their errors in the professional literature.