8,9

A chord perpendicular to the diameter of a circle divides it in the ratio $1: 3$. Under what angles is the chord seen from the ends of this diameter?
Connect one end of the chord with the ends of the diameter and the center of the circle. One of the resulting triangles is equilateral.

## Solution

Let $AB$ be the diameter of the circle with center $O$, $CD$ be the chord intersecting the diameter $AB$ at point $M$, $CD \perp AB$, and $AM: BM = 1: 3$. Then $OC = OB = OA = 2AM$, $OM = OA - AM = AM$.

Since in the right triangle $COM$, the leg $OM$ is half the hypotenuse $OC$, then $\angle OCM = 30^{\circ}$. Therefore, $\angle COM = 60^{\circ}$, and since triangle $AOC$ is equilateral, 

$\angle CAB = 60^{\circ}$. Similarly, $\angle DAB = 60^{\circ}$. Therefore, $\angle CAD = 120^{\circ}$.

From the right triangle $BAC$, we find that $\angle CBA = 30^{\circ}$. Therefore, $\angle CBD = 2 \angle CBA = 60^{\circ}$.

## Answer

$60^{\circ}, 120^{\circ}$.