Two force `F_(1)" and "F_(2)` are acting on a body. One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is :-

Options

A. `cos^(-1)(1//2)`
B. `cos^(-1)(-1//2)`
C. `cos^(-1)(-1//4)`
D. `cos^(-1)(1//4)`

Answer

To solve the problem step by step, we will analyze the forces acting on the body and apply the principles of vector addition.

### Step 1: Define the Forces
Let \( F_1 \) be one of the forces and \( F_2 \) be the other force. According to the problem, one force is double the other:
\[ F_2 = 2F_1 \]

### Step 2: Understand the Resultant Force
The resultant force \( R \) of the two forces \( F_1 \) and \( F_2 \) is given to be equal to the greater force, which is \( F_2 \):
\[ R = F_2 \]

### Step 3: Apply the Parallelogram Law of Vector Addition
The resultant of two forces can be calculated using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Substituting \( R = F_2 \) and \( F_2 = 2F_1 \) into the equation:
\[ 2F_1 = \sqrt{F_1^2 + (2F_1)^2 + 2F_1(2F_1) \cos \theta} \]

### Step 4: Simplify the Equation
Calculating \( (2F_1)^2 \):
\[ (2F_1)^2 = 4F_1^2 \]
Thus, the equation becomes:
\[ 2F_1 = \sqrt{F_1^2 + 4F_1^2 + 4F_1^2 \cos \theta} \]
This simplifies to:
\[ 2F_1 = \sqrt{5F_1^2 + 4F_1^2 \cos \theta} \]

### Step 5: Square Both Sides
Squaring both sides to eliminate the square root:
\[ (2F_1)^2 = 5F_1^2 + 4F_1^2 \cos \theta \]
This gives:
\[ 4F_1^2 = 5F_1^2 + 4F_1^2 \cos \theta \]

### Step 6: Rearranging the Equation
Rearranging the equation leads to:
\[ 4F_1^2 - 5F_1^2 = 4F_1^2 \cos \theta \]
\[ -F_1^2 = 4F_1^2 \cos \theta \]

### Step 7: Solve for \( \cos \theta \)
Dividing both sides by \( F_1^2 \) (assuming \( F_1 \neq 0 \)):
\[ -1 = 4 \cos \theta \]
Thus,
\[ \cos \theta = -\frac{1}{4} \]

### Step 8: Find the Angle \( \theta \)
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{1}{4}\right) \]

### Final Answer
The angle between the two forces is:
\[ \theta = \cos^{-1}\left(-\frac{1}{4}\right) \]