Three forces ` 9 , 12` and `15 N` acting at a point are in equilibrium . The angle between `9 N` and `15 N` is

Options

A. `cos^(-1)(3//5)`
B. `cos^(-1)(4//5)`
C. `pi - cos^(-1) (3//5)`
D. `pi - cos^(-1) (4//5)`

Explanation

To solve the problem of finding the angle between the forces of 9 N and 15 N acting at a point in equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Equilibrium**:
In equilibrium, the vector sum of all forces acting at a point must equal zero. Therefore, if we have three forces \( \vec{A} = 9 \, \text{N} \), \( \vec{B} = 12 \, \text{N} \), and \( \vec{C} = 15 \, \text{N} \), we can write:
\[ \vec{A} + \vec{B} + \vec{C} = 0 \]

2. **Rearranging the Equation**:
Rearranging gives us:
\[ \vec{A} + \vec{C} = -\vec{B} \]

3. **Using the Law of Cosines**:
To find the angle \( \theta \) between \( \vec{A} \) (9 N) and \( \vec{C} \) (15 N), we can apply the law of cosines:
\[ |\vec{A} + \vec{C}|^2 = |\vec{B}|^2 \]
Expanding this using the magnitudes and the cosine of the angle:
\[ |\vec{A}|^2 + |\vec{C}|^2 + 2 |\vec{A}| |\vec{C}| \cos \theta = |\vec{B}|^2 \]

4. **Substituting the Values**:
Substitute the magnitudes:
\[ 9^2 + 15^2 + 2 \cdot 9 \cdot 15 \cos \theta = 12^2 \]
Simplifying gives:
\[ 81 + 225 + 270 \cos \theta = 144 \]

5. **Combining Like Terms**:
Combine the constants:
\[ 306 + 270 \cos \theta = 144 \]

6. **Isolating the Cosine Term**:
Rearranging to isolate \( \cos \theta \):
\[ 270 \cos \theta = 144 - 306 \]
\[ 270 \cos \theta = -162 \]
\[ \cos \theta = -\frac{162}{270} = -\frac{3}{5} \]

7. **Finding the Angle**:
To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}(-\frac{3}{5}) \]
Since the cosine is negative, we can express this as:
\[ \theta = \pi - \cos^{-1}(\frac{3}{5}) \]

### Final Answer:
The angle between the forces of 9 N and 15 N is:
\[ \theta = \pi - \cos^{-1}\left(\frac{3}{5}\right) \]