The resultant of two forces 2 N and 3 N is `sqrt(19)` N. The angle between the forces is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between the two forces of magnitudes 2 N and 3 N, given that their resultant is \( \sqrt{19} \) N, we can use the formula for the resultant of two forces acting at an angle \( \theta \):

\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]

Where:
- \( R \) is the resultant force,
- \( F_1 \) and \( F_2 \) are the magnitudes of the two forces,
- \( \theta \) is the angle between the two forces.

### Step 1: Substitute the known values into the equation

Given:
- \( F_1 = 2 \, \text{N} \)
- \( F_2 = 3 \, \text{N} \)
- \( R = \sqrt{19} \, \text{N} \)

Substituting these values into the resultant formula:

\[ \sqrt{19} = \sqrt{2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cos \theta} \]

### Step 2: Square both sides to eliminate the square root

Squaring both sides gives:

\[ 19 = 2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cos \theta \]

Calculating \( 2^2 \) and \( 3^2 \):

\[ 19 = 4 + 9 + 12 \cos \theta \]

### Step 3: Simplify the equation

Combine the constants on the right side:

\[ 19 = 13 + 12 \cos \theta \]

### Step 4: Isolate \( \cos \theta \)

Subtract 13 from both sides:

\[ 19 - 13 = 12 \cos \theta \]

This simplifies to:

\[ 6 = 12 \cos \theta \]

### Step 5: Solve for \( \cos \theta \)

Dividing both sides by 12:

\[ \cos \theta = \frac{6}{12} = \frac{1}{2} \]

### Step 6: Find \( \theta \)

Now, we find the angle \( \theta \) using the inverse cosine function:

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

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

\[ \theta = 60^\circ \quad \text{or} \quad \theta = \frac{\pi}{3} \, \text{radians} \]

### Final Answer

The angle between the two forces is \( 60^\circ \).

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