Answer: $75^{\circ}$.

Solution. Consider triangle $A O C$, where $O$ is the center of the circle. This triangle is isosceles because $O C$ and $O A$ are radii. Therefore, by the property of isosceles triangles, angles $A$ and $C$ are equal. Draw the perpendicular $C M$ to side $A O$ and consider the right triangle $O M C$. According to the problem, the leg $C M$ is half the hypotenuse $O C$. Therefore, the measure of angle $C O M$ is $30^{\circ}$.

Then, by the theorem on the sum of the angles of a triangle, we get that angle $C A O$ (or $C A B$) is $75^{\circ}$.

## Grading Criteria.

![](https://cdn.mathpix.com/cropped/2024_05_06_3900164aa39d99b921fag-3.jpg?height=406&width=465&top_left_y=728&top_left_x=1478)

- Correct and justified solution to the problem - 7 points.
- Correct reasoning that constitutes a solution to the problem but results in an incorrect answer (e.g., angle $C O A$ is given instead of angle $C A O$) - 6 points.
- Generally correct reasoning with minor errors that do not fundamentally affect the solution, and the correct answer is given - 5 points.
- Correct solution to the problem without justifications: all intermediate conclusions are stated without indicating the connections between them (references to theorems or definitions) - 4 points.
- Additional constructions and markings on the diagram that clearly show the solution process, the correct answer is given, but the reasoning is not provided - 3 points.
- Correct answer given with incorrect reasoning - 0 points.
- Only the correct answer is given - 0 points.
5. In a circle, a diameter $A B$ and a chord $C D$ parallel to it were drawn such that the distance between them is half the radius of this circle (see figure). Find the angle $CAB$.

