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

Options

A. `7/8`
B. `3/4`
C. `4/5`
D. `3/5`

Answer

To solve the problem step by step, we will use the given information about triangle ABC where \( A = B = 120^\circ \) and \( R = 8r \). We need to find the value of \( \cos C \).

### Step-by-Step Solution:

1. **Understanding the Angles in Triangle ABC**:
Since \( A + B + C = 180^\circ \) and given \( A = B = 120^\circ \):
\[ C = 180^\circ - A - B = 180^\circ - 120^\circ - 120^\circ = -60^\circ \]
However, this is not possible in a triangle. So, we need to use the relationship between angles more carefully.

2. **Using the Relationship**:
Since \( A = B \), we can denote \( A = B = 120^\circ \). Therefore, we can express \( C \) as:
\[ C = 180^\circ - 2A = 180^\circ - 240^\circ = -60^\circ \]
This indicates a misunderstanding; we need to reconsider the relationship between the circumradius \( R \) and the inradius \( r \).

3. **Using the Relation \( R = 8r \)**:
The relationship between the circumradius \( R \) and the inradius \( r \) is given by the formula:
\[ R = \frac{abc}{4K} \quad \text{and} \quad r = \frac{K}{s} \]
where \( K \) is the area of the triangle and \( s \) is the semi-perimeter.

4. **Using the Sine Rule**:
We can also use the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
Since \( A = B \), we can write:
\[ a = b = 2R \sin(120^\circ) \quad \text{and} \quad c = 2R \sin C \]

5. **Finding \( \cos C \)**:
We know that:
\[ \cos C = -\cos(60^\circ) = -\frac{1}{2} \]
However, we need to find \( \cos C \) in terms of \( R \) and \( r \).

6. **Using the Derived Equation**:
From the derived equation in the video:
\[ 1 = 32 \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \]
Substituting \( A = B = 120^\circ \):
\[ 1 = 32 \sin(60^\circ) \sin(60^\circ) \sin\left(\frac{C}{2}\right) \]
Simplifying gives:
\[ 1 = 32 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \cdot \sin\left(\frac{C}{2}\right) \]
\[ 1 = 24 \sin\left(\frac{C}{2}\right) \]
Thus,
\[ \sin\left(\frac{C}{2}\right) = \frac{1}{24} \]

7. **Finding \( \cos C \)**:
Using the identity:
\[ \sin^2\left(\frac{C}{2}\right) + \cos^2\left(\frac{C}{2}\right) = 1 \]
\[ \cos^2\left(\frac{C}{2}\right) = 1 - \left(\frac{1}{24}\right)^2 = 1 - \frac{1}{576} = \frac{575}{576} \]
Therefore,
\[ \cos\left(\frac{C}{2}\right) = \sqrt{\frac{575}{576}} = \frac{\sqrt{575}}{24} \]
Finally, using the double angle formula:
\[ \cos C = 2\cos^2\left(\frac{C}{2}\right) - 1 = 2 \cdot \frac{575}{576} - 1 = \frac{1150 - 576}{576} = \frac{574}{576} = \frac{287}{288} \]

### Final Answer:
\[ \cos C = \frac{287}{288} \]