Given that `vec(A)+vec(B)=vec(C )`. If `|vec(A)|=4, |vec(B)|=5` and `|vec(C )|=sqrt(61)`, the angle between `vec(A)` and `vec(B)` is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Explanation

To find the angle between the vectors \(\vec{A}\) and \(\vec{B}\), we can use the formula for the magnitude of the resultant vector \(\vec{C}\) when two vectors are added. The formula is given by:

\[ |\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]

### Step 1: Identify the given values
We are given:
- \(|\vec{A}| = 4\)
- \(|\vec{B}| = 5\)
- \(|\vec{C}| = \sqrt{61}\)

### Step 2: Substitute the values into the formula
Substituting the known values into the formula, we have:

\[ \sqrt{61} = \sqrt{4^2 + 5^2 + 2 \cdot 4 \cdot 5 \cdot \cos \theta} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:

\[ 61 = 4^2 + 5^2 + 2 \cdot 4 \cdot 5 \cdot \cos \theta \]

### Step 4: Calculate \(4^2\) and \(5^2\)
Calculating the squares:

\[ 4^2 = 16 \quad \text{and} \quad 5^2 = 25 \]

So we have:

\[ 61 = 16 + 25 + 40 \cos \theta \]

### Step 5: Simplify the equation
Combining the constants:

\[ 61 = 41 + 40 \cos \theta \]

### Step 6: Isolate \(40 \cos \theta\)
Subtract 41 from both sides:

\[ 61 - 41 = 40 \cos \theta \]

This simplifies to:

\[ 20 = 40 \cos \theta \]

### Step 7: Solve for \(\cos \theta\)
Dividing both sides by 40:

\[ \cos \theta = \frac{20}{40} = \frac{1}{2} \]

### Step 8: Find the angle \(\theta\)
Now, we find the angle \(\theta\):

\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]

The angle whose cosine is \(\frac{1}{2}\) is:

\[ \theta = 60^\circ \]

### Final Answer
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(60^\circ\).

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