The sum of magnitudes of two forces acting at a point is 16 and magnitude of their resultant is `8sqrt(3)`. If the resultant is at `90^(@)` with the force of smaller magnitude, then their magnitudes are

Options

A. 3,13
B. 2 , 14
C. 5 ,11
D. 4,12

Explanation

To solve the problem, we will denote the magnitudes of the two forces as \( F_A \) and \( F_B \), where \( F_A < F_B \). We are given the following information:

1. The sum of the magnitudes of the two forces is \( F_A + F_B = 16 \).
2. The magnitude of their resultant is \( R = 8\sqrt{3} \).
3. The resultant \( R \) is at \( 90^\circ \) with the force of smaller magnitude \( F_A \).

### Step-by-step Solution:

**Step 1: Set up the equations.**

From the given information, we have:
1. \( F_A + F_B = 16 \) (Equation 1)
2. \( R = 8\sqrt{3} \) (Equation 2)

**Step 2: Use the formula for the resultant of two vectors.**

The magnitude of the resultant \( R \) of two forces \( F_A \) and \( F_B \) can be expressed as:
\[ R = \sqrt{F_A^2 + F_B^2 + 2F_A F_B \cos(\theta)} \]
Since \( R \) is at \( 90^\circ \) with \( F_A \), we know that \( \theta \) is the angle between \( F_A \) and \( F_B \). Thus, we can write:
\[ R^2 = F_A^2 + F_B^2 \]
because \( \cos(90^\circ) = 0 \).

Substituting \( R = 8\sqrt{3} \) into the equation gives:
\[ (8\sqrt{3})^2 = F_A^2 + F_B^2 \]
\[ 192 = F_A^2 + F_B^2 \quad \text{(Equation 3)} \]

**Step 3: Solve the equations simultaneously.**

Now we have two equations:
1. \( F_A + F_B = 16 \) (Equation 1)
2. \( F_A^2 + F_B^2 = 192 \) (Equation 3)

From Equation 1, we can express \( F_B \) in terms of \( F_A \):
\[ F_B = 16 - F_A \]

**Step 4: Substitute \( F_B \) into Equation 3.**

Substituting \( F_B \) into Equation 3:
\[ F_A^2 + (16 - F_A)^2 = 192 \]
Expanding the equation:
\[ F_A^2 + (256 - 32F_A + F_A^2) = 192 \]
Combining like terms:
\[ 2F_A^2 - 32F_A + 256 = 192 \]
\[ 2F_A^2 - 32F_A + 64 = 0 \]
Dividing the entire equation by 2:
\[ F_A^2 - 16F_A + 32 = 0 \]

**Step 5: Solve the quadratic equation.**

Using the quadratic formula \( F_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ F_A = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \]
\[ F_A = \frac{16 \pm \sqrt{256 - 128}}{2} \]
\[ F_A = \frac{16 \pm \sqrt{128}}{2} \]
\[ F_A = \frac{16 \pm 8\sqrt{2}}{2} \]
\[ F_A = 8 \pm 4\sqrt{2} \]

**Step 6: Find \( F_B \).**

Using \( F_B = 16 - F_A \):
1. If \( F_A = 8 + 4\sqrt{2} \), then \( F_B = 16 - (8 + 4\sqrt{2}) = 8 - 4\sqrt{2} \).
2. If \( F_A = 8 - 4\sqrt{2} \), then \( F_B = 16 - (8 - 4\sqrt{2}) = 8 + 4\sqrt{2} \).

Since \( F_A < F_B \), we take:
\[ F_A = 8 - 4\sqrt{2}, \quad F_B = 8 + 4\sqrt{2} \]

**Step 7: Calculate the numerical values.**

Calculating the approximate values:
- \( 4\sqrt{2} \approx 5.656 \)
- \( F_A \approx 8 - 5.656 \approx 2.344 \)
- \( F_B \approx 8 + 5.656 \approx 13.656 \)

Thus, the magnitudes of the forces are approximately:
\[ F_A \approx 2 \text{ N}, \quad F_B \approx 14 \text{ N} \]

### Final Answer:
The magnitudes of the two forces are \( F_A = 2 \text{ N} \) and \( F_B = 14 \text{ N} \).