In `DeltaABC`, a=5, b=7, c=8, find `cosB/2`

Explanation

To find \( \cos \frac{B}{2} \) in triangle \( ABC \) where \( a = 5 \), \( b = 7 \), and \( c = 8 \), we will follow these steps:

### Step 1: Use the Cosine Law to find \( \cos B \)

The Cosine Law states that:

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

Substituting the values of \( a \), \( b \), and \( c \):

\[ \cos B = \frac{5^2 + 8^2 - 7^2}{2 \cdot 5 \cdot 8} \]

Calculating the squares:

\[ = \frac{25 + 64 - 49}{2 \cdot 5 \cdot 8} \]

### Step 2: Simplify the numerator

Now, simplify the numerator:

\[ = \frac{25 + 64 - 49}{2 \cdot 5 \cdot 8} = \frac{40}{80} \]

### Step 3: Simplify the fraction

Now, simplify the fraction:

\[ = \frac{1}{2} \]

So, we have:

\[ \cos B = \frac{1}{2} \]

### Step 4: Use the Half-Angle Formula to find \( \cos \frac{B}{2} \)

The half-angle formula for cosine is:

\[ \cos \frac{B}{2} = \sqrt{\frac{1 + \cos B}{2}} \]

Substituting \( \cos B = \frac{1}{2} \):

\[ \cos \frac{B}{2} = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{\frac{3}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answer

Thus, the value of \( \cos \frac{B}{2} \) is:

\[ \cos \frac{B}{2} = \frac{\sqrt{3}}{2} \]

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