A circular sector is bounded by an arc of length 10 units and a chord of 8 units. Determine the area of the sector to the accuracy available from the function tables in school.
Let the center of the arc $AB$ be $O$, the radius $OA = r$, the length of the arc and the chord be $i$ and $h$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_0203be9249491cd9539ag-1.jpg?height=350&width=585&top_left_y=237&top_left_x=759)

The area $t$ of the segment is the difference between the area of the sector $AOB$ and the area of the triangle $AOB$, using the given data:

$$
t=\frac{ir}{2}-\frac{h \sqrt{r^{2}-h^{2} / 4}}{2}=5 r-4 \sqrt{r^{2}-16}
$$

Let the midpoint of the arc be $C$, and $\angle COB = x$. Then $r = \frac{h}{2 \sin x}$, and $i = 2rx$ (with $x$ in radians), from which eliminating $r$ gives:

$$
\frac{x}{\sin x}=\frac{i}{h}=\frac{5}{4}, \quad x=\frac{5}{4} \sin x
$$

We will seek an approximate solution to this equation.

A lower bound for $x$ is given by the known fact that when $x = 60^\circ = \pi / 3$, $i = r \cdot 2\pi / 3$, $h = r \sqrt{3}$, and $i / h = 2\pi / \sqrt{27} \approx 60^\circ$.

An upper bound is obtained from the right triangle $ACD$, where $D$ is the midpoint of the chord $AB$, in which $\angle DAC = x / 2$, and the hypotenuse $AC$ is smaller than half the arc:

$$
\cos \frac{x}{2}=\frac{AD}{AC}>\frac{h / 2}{i / 2}=0.8, \quad \frac{x}{2} < 1.25
$$

Thus, $x < 64.5^\circ$. The desired ratio of 1.25 lies in the interval $(1.247, 1.252)$, dividing it in the ratio $3:2$. Therefore, it is reasonable to try $64.8^\circ$ in the interval between $64.5^\circ$ and $65^\circ$. With this, both sides of (1) are 1.1310, and according to our table, the solution to (1) is $x = 64.8^\circ$.

Now, $r = \frac{h}{2 \sin x} = 4.42$ units of length, and according to the previous calculations, $t = 22.1 - 4 \sqrt{0.42 \cdot 8.42} = 14.6$ units of area.

Tóth Tibor (Szolnok, Verseghy F. g. III. o. t.) Kovács Tamás (Győr, Czuczor G. Bencés g. III. o. t.)

Remarks. 1. An initial approximate value for the equation $\frac{\sin x}{x} = 0.8$ can also be obtained by considering the first three terms of the series

$$
\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \ldots
$$

found in school function tables, solving the resulting

$$
\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} = 0.8, \quad x^4 - 20x^2 + 24 = 0
$$

and choosing the appropriate root

$$
x = \sqrt{10 - \sqrt{76}} \approx 1.132 \approx 65^\circ
$$

as the starting point.

2. The use of degree-radian tables can be avoided. Based on the relationship $x_{\text{rad}} = \left(\frac{\pi}{180^\circ}\right) \cdot x_{\text{deg}}$, (1) can be rewritten (further, let $x$ be measured in degrees):

$$
\frac{\sin x}{x} = \frac{4\pi}{900} = 0.013963 \quad(=c)
$$

For $x = 60^\circ, 70^\circ, 65^\circ, 64^\circ$, the division can be quickly performed without using logarithms, and the quotients are respectively

$$
0.01443 \quad(>c), \quad 0.01342 \quad(<c)
$$

In the last two, the ratio of the deficit and the excess is $(-0.00020):(+0.00081) = 1:4$, which again yields $64.8^\circ$.