Two equal vector have a resultant equal to either of them, then the angle between them will be:

Options

A. `110^(@)`
B. `120^(@)`
C. `60^(@)`
D. `150^(@)`

Explanation

To solve the problem of finding the angle between two equal vectors when their resultant is equal to either of them, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Vectors**: Let the two equal vectors be \( \vec{A} \) and \( \vec{A} \). Since they are equal, we can denote their magnitude as \( A \).

2. **Resultant of Two Vectors**: The formula for the resultant \( R \) of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ R = \sqrt{A^2 + A^2 + 2A \cdot A \cos \theta} \]
Here, since \( \vec{A} \) and \( \vec{B} \) are equal, we can replace \( \vec{B} \) with \( \vec{A} \).

3. **Substituting Magnitudes**: Since both vectors have the same magnitude \( A \), we can simplify the equation:
\[ R = \sqrt{A^2 + A^2 + 2A^2 \cos \theta} \]
This simplifies to:
\[ R = \sqrt{2A^2 + 2A^2 \cos \theta} \]

4. **Setting the Resultant Equal to One of the Vectors**: According to the problem, the resultant \( R \) is equal to the magnitude of either vector \( A \):
\[ A = \sqrt{2A^2 + 2A^2 \cos \theta} \]

5. **Squaring Both Sides**: To eliminate the square root, we square both sides:
\[ A^2 = 2A^2 + 2A^2 \cos \theta \]

6. **Rearranging the Equation**: Rearranging gives us:
\[ A^2 - 2A^2 = 2A^2 \cos \theta \]
Simplifying this, we have:
\[ -A^2 = 2A^2 \cos \theta \]

7. **Solving for Cosine**: Dividing both sides by \( A^2 \) (assuming \( A \neq 0 \)):
\[ -1 = 2 \cos \theta \]
Thus,
\[ \cos \theta = -\frac{1}{2} \]

8. **Finding the Angle**: The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \]

### Final Answer:
The angle between the two equal vectors is \( 120^\circ \).

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