Forces 7, 5 and 3 acting on a particle are in equilibrium, the angle between the pair of forces 5 and 3 is

Options

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

Answer

To find the angle between the forces of magnitudes 5 and 3 when they are in equilibrium with a force of magnitude 7, we can use the law of cosines. Here’s the step-by-step solution:

### Step 1: Understand the equilibrium condition
In equilibrium, the vector sum of the forces acting on a particle must be zero. This means that the resultant of the forces must balance out.

### Step 2: Set up the equation using the law of cosines
For the forces of magnitudes 5 and 3, the resultant force must equal the third force of magnitude 7. According to the law of cosines, we can express this as:
\[ R^2 = A^2 + B^2 - 2AB \cos(\theta) \]
where \( R \) is the resultant force, \( A \) and \( B \) are the magnitudes of the two forces, and \( \theta \) is the angle between them.

### Step 3: Substitute the known values
Here, \( R = 7 \), \( A = 5 \), and \( B = 3 \). Plugging these values into the equation gives us:
\[ 7^2 = 5^2 + 3^2 - 2 \cdot 5 \cdot 3 \cos(\theta) \]

### Step 4: Calculate the squares
Calculating the squares:
\[ 49 = 25 + 9 - 30 \cos(\theta) \]

### Step 5: Simplify the equation
Combine the terms on the right side:
\[ 49 = 34 - 30 \cos(\theta) \]

### Step 6: Rearrange to solve for cos(θ)
Rearranging gives:
\[ 30 \cos(\theta) = 34 - 49 \]
\[ 30 \cos(\theta) = -15 \]
\[ \cos(\theta) = -\frac{15}{30} \]
\[ \cos(\theta) = -\frac{1}{2} \]

### Step 7: Find the angle θ
The angle whose cosine is \(-\frac{1}{2}\) is:
\[ \theta = 120^\circ \]

### Final Answer
The angle between the forces of magnitudes 5 and 3 is \( 120^\circ \).

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