Resultant of two forces `vecF_(1)` and `vecF_(2)` has magnitude 50 N. The resultant is inclined to `vecF_(1)` at `60^(@)` and to `vecF_(1)` at `30^(@)` . Magnitudes of `vecF_(1)` and `vecF_(2)`, respectively, are:

Options

A. 25 N, `25sqrt(3)N`
B. 20 N, `20sqrt(3)N`
C. 20 N, 30N
D. 30 N, 40N

Answer

To solve the problem, we need to find the magnitudes of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \) given that the resultant force has a magnitude of 50 N and is inclined at angles of 60° and 30° to \( \vec{F}_1 \) and \( \vec{F}_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have two forces \( \vec{F}_1 \) and \( \vec{F}_2 \).
- The resultant \( \vec{R} \) has a magnitude of 50 N.
- The angle between \( \vec{R} \) and \( \vec{F}_1 \) is 60°.
- The angle between \( \vec{R} \) and \( \vec{F}_2 \) is 30°.

2. **Using the Cosine Rule**:
- From the geometry of the triangle formed by \( \vec{F}_1 \), \( \vec{F}_2 \), and \( \vec{R} \), we can use the cosine of the angles to find the components of the forces.
- For \( \vec{F}_1 \):
\[ \cos(60°) = \frac{F_1}{R} \]
Substituting \( R = 50 \):
\[ \cos(60°) = \frac{F_1}{50} \]
Since \( \cos(60°) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{F_1}{50} \]
Therefore, multiplying both sides by 50:
\[ F_1 = 50 \times \frac{1}{2} = 25 \text{ N} \]

3. **Finding \( \vec{F}_2 \)**:
- Now, for \( \vec{F}_2 \):
\[ \cos(30°) = \frac{F_2}{R} \]
Substituting \( R = 50 \):
\[ \cos(30°) = \frac{F_2}{50} \]
Since \( \cos(30°) = \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{F_2}{50} \]
Therefore, multiplying both sides by 50:
\[ F_2 = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ N} \]

4. **Final Result**:
- The magnitudes of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \) are:
\[ F_1 = 25 \text{ N}, \quad F_2 = 25\sqrt{3} \text{ N} \]

### Summary of Results:
- The magnitudes of \( \vec{F}_1 \) and \( \vec{F}_2 \) are \( 25 \text{ N} \) and \( 25\sqrt{3} \text{ N} \) respectively.