If `a, b, A` be given in a triangle and `c_1 and c_2` be two possible value of the third side such that `c_1^2+c_1c_2+c_2^2=a^2,` then a is equal to

Options

A. (a)`30^(@)`
B. (b)`60^(@)`
C. (c)`90^(@)`
D. (d)`120^(@)`

Answer

To solve the problem, we need to find the value of \( A \) given the relationship between the sides \( c_1 \), \( c_2 \), and \( a \) in a triangle.

### Step-by-step Solution:

1. **Understanding the Given Condition**: We are given that \( c_1^2 + c_1 c_2 + c_2^2 = a^2 \). This is a specific condition that relates the two possible lengths of the third side of the triangle to the other two sides.

2. **Using the Cosine Rule**: In any triangle, the cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cos A \]
where \( c \) is the side opposite angle \( A \), and \( a \) and \( b \) are the other two sides.

3. **Setting Up the Equations**: From the cosine rule, we can express the sum and product of the roots \( c_1 \) and \( c_2 \):
- Let \( c_1 + c_2 = 2b \cos A \) (sum of the roots)
- Let \( c_1 c_2 = b^2 - a^2 \) (product of the roots)

4. **Substituting into the Given Equation**: We can rewrite the equation \( c_1^2 + c_1 c_2 + c_2^2 \) using the identities:
\[ c_1^2 + c_2^2 = (c_1 + c_2)^2 - 2c_1 c_2 \]
Thus,
\[ c_1^2 + c_2^2 + c_1 c_2 = (c_1 + c_2)^2 - c_1 c_2 \]
Substituting the expressions for \( c_1 + c_2 \) and \( c_1 c_2 \):
\[ = (2b \cos A)^2 - (b^2 - a^2) \]
\[ = 4b^2 \cos^2 A - b^2 + a^2 \]
\[ = 3b^2 \cos^2 A + a^2 \]

5. **Equating to \( a^2 \)**: We set the above expression equal to \( a^2 \):
\[ 3b^2 \cos^2 A + a^2 = a^2 \]
This simplifies to:
\[ 3b^2 \cos^2 A = 0 \]
Since \( b^2 \) cannot be zero (as it represents a side of the triangle), we conclude that:
\[ \cos^2 A = 0 \]

6. **Finding \( A \)**: The cosine of an angle is zero at \( A = 90^\circ \) or \( A = 270^\circ \). However, since \( A \) must be an angle in a triangle, we only consider \( A = 90^\circ \).

### Conclusion:
Thus, the value of \( A \) is \( 90^\circ \).