The problem is asking to determine the radius of the circle in the diagram. The triangle $\triangle ABC$ is inscribed in a circle, with $AB = 5 \, \text{cm}$ and $\angle ACB = 30^\circ$. Here's the solution:

Step 1: Key property of an inscribed triangle

In a circle, if a triangle is inscribed, the angle subtended at the circumference by a chord (like $AB$) is half the angle subtended at the circle's center. Thus, the central angle corresponding to $\angle ACB$ is: $\angle AOB = 2 \times 30^\circ = 60^\circ$

Here, $O$ is the center of the circle.

Step 2: Use the Law of Cosines in $\triangle AOB$

In the isosceles triangle $\triangle AOB$, the sides $OA$ and $OB$ are radii ($r$) of the circle, and $AB = 5 \, \text{cm}$ is the chord. Using the Law of Cosines: $AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\angle AOB)$ Substitute the known values: $5^2 = r^2 + r^2 - 2 \cdot r \cdot r \cdot \cos(60^\circ)$ $25 = 2r^2 - r^2 \cdot 1 \quad (\text{since } \cos(60^\circ) = 0.5)$ $25 = r^2$

Step 3: Solve for $r$

$r = \sqrt{25} = 5 \, \text{cm}$

Final Answer:

The radius of the circle is: $\boxed{5 \, \text{cm}}$

Let me know if you'd like further clarification or additional explanation!

Relative Questions:

1. What is the relationship between the central and inscribed angles in a circle?
2. How does the Law of Cosines work in general triangle problems?
3. Why is $\cos(60^\circ) = 0.5$?
4. What other properties can be used to find the radius of a circle with an inscribed triangle?
5. How would the solution change if $\angle ACB$ were different?

Tip:

Always confirm if the triangle is isosceles or right-angled when working with circles, as it simplifies calculations!