| $\left[\begin{array}{l}\text { [ Right triangle with an angle of \$30^{\circ}\$ } \\ \text { [ Parallel lines, properties and criteria. Transversals ] }\end{array}\right.$ |
| :---: | :---: |
|  | $\left[\begin{array}{ll}\text { Chords and secants (other) }\end{array}\right]$ |

In a circle, a diameter $A B$ and a chord $C D$ parallel to it are drawn such that the distance between them is half the radius of the circle (see figure). Find the angle $C A B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-09.jpg?height=434&width=483&top_left_y=1301&top_left_x=793)
Let $O$ be the center of the given circle, and $K$ be the foot of the perpendicular dropped from $O$ to $CD$ (see the figure). Then, in the right triangle $OSK$, the leg $OK$ is half the hypotenuse $OC$, so $\angle OCK = 30^{\circ}$. Since $CD \parallel AB$, then $\angle AOC = \angle OCK = 30^{\circ}$. Triangle $AOC$ is isosceles, so $\angle CAO = \angle ACO = (180^{\circ} - 30^{\circ}) : 2 = 75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-09.jpg?height=423&width=483&top_left_y=2027&top_left_x=793)

## Answer

$75^{\circ}$.