12.407 A line perpendicular to the chord of a segment divides the chord in the ratio 1:4, and the arc - in the ratio $1: 2$. Find the cosine of the central angle subtended by this arc.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-291.jpg?height=255&width=363&top_left_y=623&top_left_x=163)

Let $O$ be the center of the circle with radius $R$ containing the segment $AB$, such that $\cup AN : \cup NB = 1:2$, $AM : MB = 1:4$, and $MN \perp AB$. Draw the radius $OK \perp AB$ (where $P$ is the intersection point of $OK$ and $AB$). Since the chord $AB$ and the arc $AB$ are bisected by the radius $OK$,

$MP : AP = 1.5 : 2.5 ; \cup AN : \cup NK = 1 : 0.5 ; \angle NOK = \frac{1}{6} \angle AOB$.

Drop the perpendicular $NL$ from $N$ to $OK$.

Let $\alpha = \angle NOK$, then

$\sin \alpha = \frac{NL}{ON} = \frac{MP}{R}, \sin 3\alpha = \frac{AP}{OA} = \frac{AP}{R} \Rightarrow \frac{\sin 3\alpha}{\sin \alpha} = \frac{5}{3} \Rightarrow \sin^2 \alpha = \frac{1}{3} \Rightarrow$

$\Rightarrow \cos 2\alpha = \frac{1}{3} \Rightarrow \cos \angle AOB = \cos 6\alpha = 4 \cos^3 2\alpha - 3 \cos 2\alpha = -\frac{23}{27}$.

Answer: $-\frac{23}{27}$.