there are two force each having same magnitude 10 N .one is incilined at an angle of `30^(@)` and other at an angle of `135^(@)` to the positive direction of x-axis .The x and y componenents of the resultant are

Options

A. `1.59 N hati and 12.07 Nhatj`
B. `10 N hati and 10 N hatj`
C. `1.59 N hati and 10 N hatj`
D. `1.59Nhati and 2 N hatj`

Answer

To solve the problem, we need to find the x and y components of the resultant of two forces, each with a magnitude of 10 N, inclined at angles of 30° and 135° to the positive x-axis.

### Step-by-Step Solution:

1. **Identify the Forces and Angles**:
- Force F1 has a magnitude of 10 N and is at an angle of 30°.
- Force F2 has a magnitude of 10 N and is at an angle of 135°.

2. **Calculate the Components of Each Force**:
- For F1:
- \( F_{1x} = F_1 \cos(30°) = 10 \cos(30°) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66 \, \text{N} \)
- \( F_{1y} = F_1 \sin(30°) = 10 \sin(30°) = 10 \times \frac{1}{2} = 5 \, \text{N} \)

- For F2:
- \( F_{2x} = F_2 \cos(135°) = 10 \cos(135°) = 10 \times -\frac{\sqrt{2}}{2} = -5\sqrt{2} \approx -7.07 \, \text{N} \)
- \( F_{2y} = F_2 \sin(135°) = 10 \sin(135°) = 10 \times \frac{\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07 \, \text{N} \)

3. **Sum the Components**:
- Total x-component of the resultant \( R_x \):
\[ R_x = F_{1x} + F_{2x} = 8.66 + (-7.07) = 1.59 \, \text{N} \]
- Total y-component of the resultant \( R_y \):
\[ R_y = F_{1y} + F_{2y} = 5 + 7.07 = 12.07 \, \text{N} \]

4. **Resultant Components**:
- The x-component of the resultant is approximately \( R_x \approx 1.59 \, \text{N} \).
- The y-component of the resultant is approximately \( R_y \approx 12.07 \, \text{N} \).

### Final Result:
The x and y components of the resultant force are approximately:
- \( R_x \approx 1.59 \, \text{N} \)
- \( R_y \approx 12.07 \, \text{N} \)