Two forces P and 2P are acting at an angle `theta`. If thesquare of the resultant of them is equal to half of the sum of the squares of their magnitude then `theta` is

Options

A. `sin^(-1) ((5)/(8))`
B. `cos^(-1) ((-5)/(8))`
C. `cos^(-1) ((8)/(5))`
D. `0^(@)`

Answer

To solve the problem, we need to find the angle \( \theta \) between two forces \( P \) and \( 2P \) given that the square of the resultant of these forces is equal to half of the sum of the squares of their magnitudes.

### Step-by-Step Solution:

1. **Identify the Forces and Their Magnitudes**:
- Let the first force be \( \vec{F_1} = P \).
- Let the second force be \( \vec{F_2} = 2P \).
- The angle between the two forces is \( \theta \).

2. **Calculate the Magnitude of the Resultant Force**:
The magnitude of the resultant \( R \) of two vectors can be calculated using the formula:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos(\theta)} \]
Substituting the values of \( F_1 \) and \( F_2 \):
\[ R = \sqrt{P^2 + (2P)^2 + 2 \cdot P \cdot 2P \cdot \cos(\theta)} \]
\[ R = \sqrt{P^2 + 4P^2 + 4P^2 \cos(\theta)} \]
\[ R = \sqrt{5P^2 + 4P^2 \cos(\theta)} \]

3. **Square the Resultant**:
We need to find \( R^2 \):
\[ R^2 = 5P^2 + 4P^2 \cos(\theta) \]

4. **Calculate Half the Sum of the Squares of the Magnitudes**:
The sum of the squares of the magnitudes of the forces is:
\[ P^2 + (2P)^2 = P^2 + 4P^2 = 5P^2 \]
Half of this sum is:
\[ \frac{1}{2}(5P^2) = \frac{5P^2}{2} \]

5. **Set Up the Equation**:
According to the problem, the square of the resultant is equal to half of the sum of the squares of their magnitudes:
\[ 5P^2 + 4P^2 \cos(\theta) = \frac{5P^2}{2} \]

6. **Simplify the Equation**:
Rearranging the equation gives:
\[ 4P^2 \cos(\theta) = \frac{5P^2}{2} - 5P^2 \]
\[ 4P^2 \cos(\theta) = \frac{5P^2}{2} - \frac{10P^2}{2} \]
\[ 4P^2 \cos(\theta) = -\frac{5P^2}{2} \]

7. **Divide by \( P^2 \)** (assuming \( P \neq 0 \)):
\[ 4 \cos(\theta) = -\frac{5}{2} \]

8. **Solve for \( \cos(\theta) \)**:
\[ \cos(\theta) = -\frac{5}{8} \]

9. **Find \( \theta \)**:
To find \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}\left(-\frac{5}{8}\right) \]

### Final Answer:
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(-\frac{5}{8}\right) \]