In `DeltaABC`, if a=6, b=8, c=10, evaluate cos2A.

Explanation

To evaluate \( \cos 2A \) in triangle \( ABC \) where \( a = 6 \), \( b = 8 \), and \( c = 10 \), we can follow these steps:

### Step 1: Use the Law of Cosines to find \( \cos A \)

The Law of Cosines states:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

Substituting the values:
- \( a = 6 \)
- \( b = 8 \)
- \( c = 10 \)

We have:
\[ \cos A = \frac{8^2 + 10^2 - 6^2}{2 \cdot 8 \cdot 10} \]

Calculating the squares:
\[ 8^2 = 64, \quad 10^2 = 100, \quad 6^2 = 36 \]

Now substituting these values:
\[ \cos A = \frac{64 + 100 - 36}{2 \cdot 8 \cdot 10} = \frac{128}{160} \]

### Step 2: Simplify \( \cos A \)

Now simplifying \( \frac{128}{160} \):
\[ \cos A = \frac{128 \div 32}{160 \div 32} = \frac{4}{5} \]

### Step 3: Use the double angle formula to find \( \cos 2A \)

The double angle formula for cosine is:
\[ \cos 2A = 2 \cos^2 A - 1 \]

Substituting \( \cos A = \frac{4}{5} \):
\[ \cos 2A = 2 \left( \frac{4}{5} \right)^2 - 1 \]

Calculating \( \left( \frac{4}{5} \right)^2 \):
\[ \left( \frac{4}{5} \right)^2 = \frac{16}{25} \]

Now substituting back:
\[ \cos 2A = 2 \cdot \frac{16}{25} - 1 = \frac{32}{25} - 1 \]

### Step 4: Convert 1 to a fraction and simplify

Convert 1 to a fraction with a denominator of 25:
\[ 1 = \frac{25}{25} \]

Now substituting:
\[ \cos 2A = \frac{32}{25} - \frac{25}{25} = \frac{32 - 25}{25} = \frac{7}{25} \]

### Final Answer
The value of \( \cos 2A \) is:
\[ \cos 2A = \frac{7}{25} \]

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