A vector of magnitude 30 and direction eastwards is added with another vector of magnitude 40 and direction Northwards . Find the magnitude and direction of resultant with the east.

Options

A. `53,53^(@)` with East
B. `50,53^(@)` with West
C. `50,53^(@)` with East
D. `53,53^(@)` with West

Answer

To solve the problem of finding the magnitude and direction of the resultant vector when a vector of magnitude 30 directed east is added to another vector of magnitude 40 directed north, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Let vector A be the eastward vector with a magnitude of 30 units.
- Let vector B be the northward vector with a magnitude of 40 units.

2. **Draw the Vectors**:
- Draw vector A (30 units) pointing to the right (east).
- Draw vector B (40 units) pointing upwards (north).
- These vectors form a right triangle where vector A is the base and vector B is the height.

3. **Calculate the Magnitude of the Resultant Vector**:
- The magnitude of the resultant vector \( R \) can be calculated using the Pythagorean theorem:
\[ R = \sqrt{A^2 + B^2} \]
- Substitute the values:
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ units} \]

4. **Calculate the Direction of the Resultant Vector**:
- To find the angle \( \theta \) that the resultant vector makes with the east direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{B}{A} = \frac{40}{30} \]
- Simplifying gives:
\[ \tan(\theta) = \frac{4}{3} \]
- Now, we find \( \theta \) using the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
- Using a calculator, we find:
\[ \theta \approx 53.13^\circ \]

5. **Final Result**:
- The magnitude of the resultant vector is 50 units.
- The direction of the resultant vector is approximately 53 degrees from the east towards the north.

### Summary:
- **Magnitude**: 50 units
- **Direction**: 53 degrees from east