If in ` A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7` then `a^2+b^2+c^2` must be `R^2` (b) `3R^2` (c) `4R^2` (d) `7R^2`

Options

A. `R^(2)`
B. `3R^(2)`
C. `4R^(2)`
D. `7R^(2)`

Explanation

To solve the problem, we need to find the value of \( a^2 + b^2 + c^2 \) in triangle \( ABC \) where the angles are given as \( A = \frac{\pi}{7} \), \( B = \frac{2\pi}{7} \), and \( C = \frac{4\pi}{7} \). We will use the sine rule and properties of triangles to reach the solution.

### Step-by-Step Solution:

1. **Use the Sine Rule**:
According to the sine rule, we have:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
where \( R \) is the circumradius of triangle \( ABC \).

2. **Substituting Values**:
Substitute the expressions for \( a \), \( b \), and \( c \):
\[ a^2 + b^2 + c^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2 \]
This simplifies to:
\[ a^2 + b^2 + c^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C) \]

3. **Using the Identity for Sine Squared**:
We can express \( \sin^2 A \), \( \sin^2 B \), and \( \sin^2 C \) using the identity:
\[ \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \]
Thus,
\[ \sin^2 A = \frac{1 - \cos(2A)}{2}, \quad \sin^2 B = \frac{1 - \cos(2B)}{2}, \quad \sin^2 C = \frac{1 - \cos(2C)}{2} \]

4. **Calculating \( \sin^2 A + \sin^2 B + \sin^2 C \)**:
Therefore,
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{3}{2} - \frac{1}{2}(\cos(2A) + \cos(2B) + \cos(2C)) \]

5. **Finding \( \cos(2A) + \cos(2B) + \cos(2C) \)**:
We need to calculate \( \cos(2A) + \cos(2B) + \cos(2C) \):
- \( 2A = \frac{2\pi}{7} \)
- \( 2B = \frac{4\pi}{7} \)
- \( 2C = \frac{8\pi}{7} \)

We can use the property of cosine:
\[ \cos(2A) + \cos(2B) + \cos(2C) = \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{8\pi}{7}\right) \]
It is known that:
\[ \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{8\pi}{7}\right) = -\frac{1}{2} \]

6. **Substituting Back**:
Now substituting back into our equation:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{3}{2} - \frac{1}{2}\left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{4} = \frac{6 + 1}{4} = \frac{7}{4} \]

7. **Final Calculation**:
Now substituting this into our equation for \( a^2 + b^2 + c^2 \):
\[ a^2 + b^2 + c^2 = 4R^2 \cdot \frac{7}{4} = 7R^2 \]

### Conclusion:
Thus, the value of \( a^2 + b^2 + c^2 \) is \( 7R^2 \).

### Answer:
The correct option is (d) \( 7R^2 \).