According to the school's four-digit function table,

$$
\sin 60^{\circ}-\sin 50^{\circ}=0.1000
$$

According to this, the angle $x$ for which

$$
\sin x=\frac{\sqrt{3}}{2}-\frac{1}{10}
$$

holds exactly is close to $50^{\circ}$. Determine whether $x$ is smaller or larger than $50^{\circ}$: (No other function table may be used.)
For the triple of a $50^{\circ}$ angle, which is $150^{\circ}$, we know the trigonometric functions precisely, so let's form one of the functions of the nearby $3x$, for example, its sine. Based on the identity $\sin 3 x=3 \sin x-4 \sin ^{3} x$,

$$
\sin 3 x=\frac{3}{10}(5 \sqrt{3}-1)-\frac{4}{1000}(390 \sqrt{3}-226)=\frac{151-15 \sqrt{3}}{250}
$$

on the other hand, $\sin 150^{\circ}=1 / 2$. The difference between the two is

$$
\frac{150-15 \sqrt{3}}{250}-\frac{1}{2}=\frac{26-15 \sqrt{3}}{250}=\frac{\sqrt{676}-\sqrt{675}}{250}>0 .
$$

Accordingly,

$$
\sin 3 x>\sin 150^{\circ}
$$

and since the function $y=\sin x$ is decreasing in the interval $\left(90^{\circ}, 270^{\circ}\right)$, which includes $150^{\circ}$, it takes a larger value at a smaller place, so $3 x$ is less than $150^{\circ}$, and $x$ is less than $50^{\circ}$.

Zoltán Bódi (Makó, József A. g. III. o. t.)