Task B-2.5. $\overline{A B}$ and $\overline{C D}$ are two mutually perpendicular chords of a circle with a radius of 10 cm. Two diameters of this circle, one through point $A$ and the other through point $B$, divide the chord $\overline{C D}$ into three equal parts. Determine the sine of the smaller central angle subtended by the chord $\overline{C D}$, if the length of the chord $\overline{A B}$ is 16 cm.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_2595713ee0b414c2653fg-09.jpg?height=754&width=705&top_left_y=1103&top_left_x=683)

$$
\begin{aligned}
& |C K|=|K L|=|L D|=\frac{1}{3}|C D| \\
& |K F|=\frac{1}{2}|K L|=\frac{1}{2} \cdot \frac{1}{3}|C D|=\frac{1}{6}|C D|
\end{aligned}
$$

Angle $B A E$ is a right angle because it subtends the diameter, so

$$
|A E|=\sqrt{|B E|^{2}-|A B|^{2}}=12 \text{ cm}
$$

From the similarity of triangles $A B E$ and $F S K$, we have

$$
\frac{|A B|}{|A E|}=\frac{|S F|}{|K F|}, \quad \frac{16}{12}=\frac{|S F|}{\frac{1}{6}|C D|}, \quad|S F|=\frac{2}{9}|C D|
$$

From the right triangle $D F S$, we get

$$
\begin{aligned}
\left(\frac{2}{9}|C D|\right)^{2}+\left(\frac{|C D|}{2}\right)^{2} & =100 \\
\frac{97}{324}|C D|^{2} & =100 \\
|C D|=\frac{180}{\sqrt{97}} & =\frac{180 \sqrt{97}}{97} \text{ cm}
\end{aligned}
$$

The inscribed angle $\alpha$, subtending the chord $\overline{C D}$, is equal to half of the corresponding central angle $\angle C S D$, that is

$$
\alpha=\frac{1}{2} \angle C S D=\angle F S D
$$

Then

$$
\sin \alpha=\frac{\frac{1}{2}|C D|}{r}=\frac{9 \sqrt{97}}{97}
$$

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant  Poreč, April 26, 2012.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.