In `DeltaABC, a=9, b=8` and c=4, then `3cosB-6cosC=?`

Answer

To solve the problem, we need to find the value of \(3 \cos B - 6 \cos C\) in triangle \(ABC\) where \(a = 9\), \(b = 8\), and \(c = 4\). We will use the cosine rule to find \(\cos B\) and \(\cos C\) first.

### Step 1: Calculate \(\cos B\)

Using the cosine rule:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

Substituting the values:
- \(a = 9\)
- \(b = 8\)
- \(c = 4\)

We have:
\[ \cos B = \frac{9^2 + 4^2 - 8^2}{2 \cdot 9 \cdot 4} \]
Calculating the squares:
\[ = \frac{81 + 16 - 64}{72} \]
\[ = \frac{33}{72} \]

### Step 2: Calculate \(\cos C\)

Using the cosine rule:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

Substituting the values:
\[ \cos C = \frac{9^2 + 8^2 - 4^2}{2 \cdot 9 \cdot 8} \]
Calculating the squares:
\[ = \frac{81 + 64 - 16}{144} \]
\[ = \frac{129}{144} \]

### Step 3: Substitute \(\cos B\) and \(\cos C\) into the expression

Now we substitute \(\cos B\) and \(\cos C\) into the expression \(3 \cos B - 6 \cos C\):
\[ 3 \cos B - 6 \cos C = 3 \left(\frac{33}{72}\right) - 6 \left(\frac{129}{144}\right) \]

Calculating \(3 \cos B\):
\[ 3 \cos B = \frac{99}{72} \]

Calculating \(6 \cos C\):
\[ 6 \cos C = \frac{774}{144} = \frac{387}{72} \]

### Step 4: Combine the results

Now we can combine the two results:
\[ 3 \cos B - 6 \cos C = \frac{99}{72} - \frac{387}{72} \]
\[ = \frac{99 - 387}{72} = \frac{-288}{72} = -4 \]

### Final Answer
Thus, the value of \(3 \cos B - 6 \cos C\) is \(-4\).

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