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In the given image, the problem involves a triangle \( \triangle ABC \) with the following details:

- \( AB = 2 \), 
- \( BC = 6 \),
- \( \angle ABC = 60^\circ \),
- and the task is to find the length of \( AP \), where \( P \) is a point of tangency between the incircle and side \( AB \).

The hint mentions that there are at least two methods to solve this problem. Let's break down how to approach it:

### Method 1: Using the Incenter and Incircle Properties
In a triangle, the incenter is the point where the angle bisectors meet, and it is the center of the incircle. The length of the tangent from any vertex to the point of tangency is the semiperimeter minus the opposite side.

1. **Calculate the semi-perimeter \( s \):**
   \[
   s = \frac{AB + BC + AC}{2}
   \]
   Since \( AB = 2 \), \( BC = 6 \), we need to find \( AC \) using the Law of Cosines.

2. **Apply the Law of Cosines:**
   \[
   AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(60^\circ)
   \]
   With \( \cos(60^\circ) = 0.5 \), the calculation simplifies to:
   \[
   AC^2 = 2^2 + 6^2 - 2 \times 2 \times 6 \times 0.5
   \]
   \[
   AC^2 = 4 + 36 - 12 = 28 \quad \Rightarrow \quad AC = \sqrt{28} = 2\sqrt{7}
   \]

3. **Find the semi-perimeter:**
   \[
   s = \frac{2 + 6 + 2\sqrt{7}}{2} = 4 + \sqrt{7}
   \]

4. **Find the length of \( AP \):**
   The length of \( AP \) is given by:
   \[
   AP = s - BC = (4 + \sqrt{7}) - 6 = \sqrt{7} - 2
   \]

### Method 2: Use of Trigonometry and Incircle Radius
The radius \( r \) of the incircle can also be determined using the area of the triangle and its semi-perimeter. The area \( A \) of the triangle can be found using:

1. **Calculate the area \( A \) using trigonometry:**
   \[
   A = \frac{1}{2} \times AB \times BC \times \sin(60^\circ)
   \]
   Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we have:
   \[
   A = \frac{1}{2} \times 2 \times 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}
   \]

2. **Use the area to find the incircle radius \( r \):**
   \[
   A = s \times r \quad \Rightarrow \quad r = \frac{A}{s} = \frac{3\sqrt{3}}{4 + \sqrt{7}}
   \]
   Once \( r \) is found, it can be used to determine the distances from the vertices to the points of tangency.

These are the steps to approach the problem. Would you like further clarification or details on any part of the solution? Here are some related questions:

1. How is the semi-perimeter used in relation to the incircle?
2. Why does the Law of Cosines work for finding side \( AC \)?
3. What are the properties of the incenter that make it helpful for this problem?
4. How can trigonometry help in finding the area of a triangle?
5. What role does the incircle radius play in the solution?

**Tip:** Understanding triangle properties, such as the Law of Cosines and the relationship between the incenter and incircle, can significantly simplify complex geometric problems.

線AB=2, BC=6, ∠ABC=60°とするとき、APの長さを求めよ。

Learn how to solve a triangle geometry problem where AB=2, BC=6, and ∠ABC=60°. The solution uses the Law of Cosines and incircle properties to find the length of AP. Suitable for high school students in Grades 10-12.

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