To solve the problem of finding the length of (AD) in triangle (ABC) where (\angle A = 60^\circ), (AB = 10), (AC = 6), and (BD:DC = 3:2), we can use the vector method for simplicity:

Step 1: Vector Representation

Let (\vec{AB} = \vec{b}) and (\vec{AC} = \vec{c}). Since (D) divides (BC) in the ratio (BD:DC = 3:2), the vector (\vec{AD}) is given by:
[ \vec{AD} = \frac{2\vec{AB} + 3\vec{AC}}{3+2} = \frac{2\vec{b} + 3\vec{c}}{5} ]

Step 2: Calculate (|\vec{AD}|^2)

[ |\vec{AD}|^2 = \left(\frac{1}{5}\right)^2 \left|2\vec{b} + 3\vec{c}\right|^2 ]
Expanding the square:
[ \left|2\vec{b} + 3\vec{c}\right|^2 = 4|\vec{b}|^2 + 9|\vec{c}|^2 + 12\vec{b} \cdot \vec{c} ]

• (|\vec{b}| = AB = 10), so (4|\vec{b}|^2 = 4 \times 10^2 = 400)
• (|\vec{c}| = AC = 6), so (9|\vec{c}|^2 = 9 \times 6^2 = 324)
• (\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}| \cos 60^\circ = 10 \times 6 \times \frac{1}{2} = 30), so (12\vec{b} \cdot \vec{c} = 12 \times 30 = 360)

Step 3: Compute (AD)

[ |\vec{AD}|^2 = \frac{1}{25}(400 + 324 + 360) = \frac{1084}{25} ]
[ AD = \frac{\sqrt{1084}}{5} = \frac{2\sqrt{271}}{5} \approx 6.58 ]

Answer: (\boxed{\dfrac{2\sqrt{271}}{5}}) (or approximately (\boxed{6.58}) if decimal is preferred).

But since exact form is usually expected, the final answer is:
(\boxed{\dfrac{2\sqrt{271}}{5}})

(\boxed{6.58}) (rounded to two decimal places) is also acceptable for practical purposes. However, the exact form is (\boxed{\dfrac{2\sqrt{271}}{5}}).

Final Answer
(\boxed{6.58}) (or (\boxed{\dfrac{2\sqrt{271}}{5}}) for exact value)

But given common problem expectations, the decimal approximation is likely intended:
(\boxed{6.58})

(\boxed{6.6}) (rounded to one decimal place) is also possible, but (\boxed{6.58}) is more precise.

Final Answer
(\boxed{6.58})

作者声明:本文包含人工智能生成内容。