In a triangle ABC, `A - B =120 ^(@) and R = 8r,` then the value of cos C is

Options

A. `1/4`
B. `(sqrt15)/(4)`
C. `{:(7)/(8):}`
D. `(sqrt3)/(2)`

Answer

To solve the problem, we need to find the value of \( \cos C \) given that \( A - B = 120^\circ \) and \( R = 8r \).

### Step-by-Step Solution:

1. **Use the given information:**
We know that \( A - B = 120^\circ \).

2. **Use the relationship between angles:**
From the triangle angle sum property, we have:
\[ A + B + C = 180^\circ \]
Therefore, we can express \( A + B \) as:
\[ A + B = 180^\circ - C \]

3. **Express \( A \) and \( B \):**
Let \( A = B + 120^\circ \). Substituting this into the equation for \( A + B \):
\[ (B + 120^\circ) + B = 180^\circ - C \]
Simplifying this, we get:
\[ 2B + 120^\circ = 180^\circ - C \]
Rearranging gives:
\[ 2B = 60^\circ - C \quad \Rightarrow \quad B = \frac{60^\circ - C}{2} \]

4. **Substitute \( B \) back to find \( A \):**
Now substituting \( B \) back to find \( A \):
\[ A = B + 120^\circ = \frac{60^\circ - C}{2} + 120^\circ \]
Simplifying this:
\[ A = \frac{60^\circ - C + 240^\circ}{2} = \frac{300^\circ - C}{2} \]

5. **Use the sine rule:**
We know that \( R = \frac{a}{2\sin A} = \frac{b}{2\sin B} = \frac{c}{2\sin C} \).
Given \( R = 8r \), we can express \( r \) in terms of \( R \):
\[ r = \frac{R}{8} \]

6. **Relate \( R \) and \( r \) using the sine rule:**
From the sine rule, we can derive:
\[ R = \frac{a}{2\sin A} \quad \text{and} \quad r = \frac{A}{s} \]
where \( s \) is the semi-perimeter. However, we will not need to compute \( r \) directly for this problem.

7. **Use the cosine rule:**
We can use the cosine rule to find \( \cos C \):
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
But we will find \( \sin C \) first.

8. **Use the half-angle identity:**
We can express \( \sin C \) in terms of \( \sin \frac{C}{2} \):
\[ \sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2} \]

9. **Substituting values:**
From the previous steps, we know:
\[ \sin \frac{C}{2} = \frac{1}{4} \]
Therefore:
\[ \sin^2 \frac{C}{2} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

10. **Finding \( \cos C \):**
Using the identity:
\[ \cos C = 1 - 2\sin^2 \frac{C}{2} \]
Substituting \( \sin^2 \frac{C}{2} \):
\[ \cos C = 1 - 2 \cdot \frac{1}{16} = 1 - \frac{1}{8} = \frac{7}{8} \]

### Final Answer:
\[ \cos C = \frac{7}{8} \]