given that `vecA+vecB+vecC=0` out of three vectors two are equal in magnitude and the magnitude of third vector is `sqrt2` times that of either of the two having equal magnitude. Then the angles between vectors are given by

Options

A. `30^@`,`60^@`,`90^@`
B. `40^@`,`45^@`,90^@`
C. `45^@`,`60^@`,`90^@`
D. `90^@`,`135^@`,`135^@`

Explanation

To solve the problem, we need to analyze the given vectors and their relationships based on the information provided.

### Step-by-Step Solution:

1. **Understanding the Vector Equation**:
We are given that \(\vec{A} + \vec{B} + \vec{C} = 0\). This implies that the vectors form a closed triangle when placed head to tail.

2. **Setting Magnitudes**:
Let the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) be \(x\). According to the problem, the magnitude of vector \(\vec{C}\) is \(\sqrt{2}\) times that of either \(\vec{A}\) or \(\vec{B}\). Therefore, we can write:
\[ |\vec{C}| = \sqrt{2}x \]

3. **Forming a Right Triangle**:
Since the vectors sum to zero, they can be represented as a triangle. Given that \(|\vec{C}| = \sqrt{2}x\) and \(|\vec{A}| = | \vec{B}| = x\), we can visualize this as a right triangle where \(\vec{C}\) is the hypotenuse.

4. **Applying the Pythagorean Theorem**:
In a right triangle, the relationship between the sides is given by:
\[ c^2 = a^2 + b^2 \]
Here, \(c = |\vec{C}| = \sqrt{2}x\), and \(a = |\vec{A}| = x\), \(b = |\vec{B}| = x\). Thus:
\[ (\sqrt{2}x)^2 = x^2 + x^2 \]
Simplifying this gives:
\[ 2x^2 = 2x^2 \]
This confirms that the triangle is indeed a right triangle.

5. **Finding Angles**:
- The angle \(\beta\) between \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\) because they are the two sides of the right triangle.
- To find the angles \(\alpha\) (between \(\vec{B}\) and \(\vec{C}\)) and \(\gamma\) (between \(\vec{A}\) and \(\vec{C}\)), we can use the properties of a right triangle:
- Since \(\vec{C}\) is the hypotenuse, the angles opposite the sides of equal lengths (both \(x\)) will be equal. Therefore, \(\alpha = \gamma\).
- The sum of angles in a triangle is \(180^\circ\):
\[ \alpha + \beta + \gamma = 180^\circ \]
Substituting \(\beta = 90^\circ\):
\[ \alpha + \gamma = 90^\circ \]
Since \(\alpha = \gamma\), we can write:
\[ 2\alpha = 90^\circ \implies \alpha = 45^\circ \quad \text{and} \quad \gamma = 45^\circ \]

6. **Final Angles**:
- \(\angle A = 90^\circ\)
- \(\angle B = 45^\circ\)
- \(\angle C = 45^\circ\)

Thus, the angles between the vectors are:
- \(\angle A = 90^\circ\)
- \(\angle B = 45^\circ\)
- \(\angle C = 45^\circ\)