A displacment vector fo magnitude 4 makes an angle `30^(@)` with the x-axis. Its rectangular components in x-y plane are

Options

A. `2sqrt(3),2`
B. `4sqrt(3),4`
C. `(2)/(sqrt(3)),2`
D. `(4)/(sqrt(3)),4`

Answer

To find the rectangular components of a displacement vector with a given magnitude and angle, we can use trigonometric functions. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Magnitude of the displacement vector (S) = 4
- Angle with the x-axis (θ) = 30 degrees

### Step 2: Calculate the x-component
The x-component of the displacement vector can be calculated using the cosine function:
\[ S_x = S \cdot \cos(\theta) \]
Substituting the known values:
\[ S_x = 4 \cdot \cos(30^\circ) \]
Using the value of \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
\[ S_x = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

### Step 3: Calculate the y-component
The y-component of the displacement vector can be calculated using the sine function:
\[ S_y = S \cdot \sin(\theta) \]
Substituting the known values:
\[ S_y = 4 \cdot \sin(30^\circ) \]
Using the value of \(\sin(30^\circ) = \frac{1}{2}\):
\[ S_y = 4 \cdot \frac{1}{2} = 2 \]

### Step 4: State the final answer
The rectangular components of the displacement vector are:
- x-component: \(2\sqrt{3}\)
- y-component: \(2\)

Thus, the answer is:
\[ \text{Rectangular components: } (2\sqrt{3}, 2) \]

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