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This problem involves triangle ABC with the following details:
- \( AB = 8 \),
- \( AC = 5 \),
- \( \angle BAC = 60^\circ \),
and we are also concerned with the circumcircle (外接円) of this triangle, denoted as \( K \).

Let's solve it step by step:

### (1) Finding \( BC \)
Using the **Law of Cosines** for \( \triangle ABC \):
\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)
\]
Substitute the given values:
\[
BC^2 = 8^2 + 5^2 - 2 \cdot 8 \cdot 5 \cdot \cos(60^\circ)
\]
Since \( \cos(60^\circ) = \frac{1}{2} \), this simplifies to:
\[
BC^2 = 64 + 25 - 80 \cdot \frac{1}{2} = 64 + 25 - 40 = 49
\]
\[
BC = \sqrt{49} = 7
\]
Thus, \( BC = 7 \).

### (2) Finding the radius \( r \) of the circumcircle \( K \)
The formula for the radius \( r \) of the circumcircle is:
\[
r = \frac{ABC}{s}
\]
where \( ABC \) is the area of the triangle and \( s \) is the semi-perimeter.

First, calculate the semi-perimeter \( s \):
\[
s = \frac{AB + AC + BC}{2} = \frac{8 + 5 + 7}{2} = 10
\]

Now, calculate the area of the triangle \( ABC \). Using the formula for the area based on two sides and the included angle:
\[
\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin(\angle BAC)
\]
Substitute the values:
\[
\text{Area} = \frac{1}{2} \cdot 8 \cdot 5 \cdot \sin(60^\circ) = \frac{1}{2} \cdot 8 \cdot 5 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}
\]

Finally, the radius \( r \) is:
\[
r = \frac{AB \cdot AC \cdot BC}{4 \cdot \text{Area}} = \frac{8 \cdot 5 \cdot 7}{4 \cdot 10\sqrt{3}} = \frac{280}{40\sqrt{3}} = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}
\]
Thus, \( K \)'s radius is \( \frac{7\sqrt{3}}{3} \).

### (3) Determining the length of segment \( CP \)
The problem involves placing point \( P \) on \( BC \) such that \( BP = 3 \). We are asked to find the length of segment \( CP \).

Since \( BC = 7 \) and \( BP = 3 \), we have:
\[
CP = BC - BP = 7 - 3 = 4
\]

### Final Answers:
1. \( BC = 7 \)
2. \( K \)'s radius is \( \frac{7\sqrt{3}}{3} \)
3. \( CP = 4 \)

Would you like further details or explanations on any part?

Here are 5 related questions to explore further:
1. What is the general form of the Law of Cosines and how can it be applied in other problems?
2. How is the circumradius related to the area of a triangle?
3. Can we derive the circumradius for a right triangle without using trigonometry?
4. What is the significance of the semi-perimeter in triangle geometry?
5. How can we apply Heron’s formula to find the area of \( \triangle ABC \) instead?

**Tip:** Whenever using trigonometric formulas, be mindful of angle units (degrees vs. radians) to avoid errors in calculations.

三角形ABCがあり, AB=8, AC=5, ∠BAC=60°である。この三角形の外接円をKとする。(1) BC=□である。(2) Kの半径は□である。(3) KのAを含まない方の弧BC上に, 点PをBP=3を満たすようにとる。このとき, 線分CPの長さは□である。

Solve a triangle ABC problem with AB=8, AC=5, and ∠BAC=60°. Learn how to find the length of BC, the radius of the circumcircle K, and the segment length CP when BP=3. This problem covers trigonometry, triangle geometry, and the Law of Cosines.

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