Two vectors `vec(a)` and `vec(b)` are at an angle of `60^(@)` with each other . Their resultant makes an angle of `45^(@)` with `vec(a)` If `|vec(b)|=2`unit , then `|vec(a)|` is

Options

A. `sqrt(3)`
B. `sqrt(3)-1`
C. `sqrt(3)+1`
D. `sqrt(3)//2`

Explanation

To solve the problem step by step, we will use the given information about the vectors and apply the relevant formulas.

### Step 1: Identify the Given Information
We have two vectors, \( \vec{A} \) and \( \vec{B} \):
- The angle between \( \vec{A} \) and \( \vec{B} \) is \( \theta = 60^\circ \).
- The magnitude of \( \vec{B} \) is \( |\vec{B}| = 2 \) units.
- The angle between the resultant vector \( \vec{R} \) and \( \vec{A} \) is \( \phi = 45^\circ \).

### Step 2: Use the Formula for the Tangent of the Angle
We can use the formula for the tangent of the angle between the resultant vector and one of the vectors:
\[ \tan(\phi) = \frac{|\vec{B}| \sin(\theta)}{|\vec{A}| + |\vec{B}| \cos(\theta)} \]
Substituting the known values:
- \( \tan(45^\circ) = 1 \)
- \( |\vec{B}| = 2 \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(60^\circ) = \frac{1}{2} \)

### Step 3: Substitute the Values into the Formula
Now we substitute these values into the formula:
\[ 1 = \frac{2 \cdot \frac{\sqrt{3}}{2}}{|\vec{A}| + 2 \cdot \frac{1}{2}} \]
This simplifies to:
\[ 1 = \frac{\sqrt{3}}{|\vec{A}| + 1} \]

### Step 4: Cross-Multiply to Solve for \( |\vec{A}| \)
Cross-multiplying gives us:
\[ |\vec{A}| + 1 = \sqrt{3} \]

### Step 5: Isolate \( |\vec{A}| \)
Now, we isolate \( |\vec{A}| \):
\[ |\vec{A}| = \sqrt{3} - 1 \]

### Step 6: Conclusion
Thus, the magnitude of vector \( \vec{A} \) is:
\[ |\vec{A}| = \sqrt{3} - 1 \]

### Final Answer
The magnitude of \( \vec{A} \) is \( \sqrt{3} - 1 \) units.

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