A resultant of two vectors makes ` 30^(@)` with one vector and ` 45^(@)` with the other. Find the two vectors if the resultant has the magnitude 15

Answer

To solve the problem, we need to find the magnitudes of two vectors \( P \) and \( Q \) given that their resultant \( R \) has a magnitude of 15 and makes angles of \( 30^\circ \) with vector \( P \) and \( 45^\circ \) with vector \( Q \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Magnitude of the resultant vector \( R = 15 \)
- Angle with vector \( P \) (let's denote it as \( \alpha = 30^\circ \))
- Angle with vector \( Q \) (let's denote it as \( \beta = 45^\circ \))

2. **Use the Cosine Rule to Find Vector \( P \):**
- The formula for the component of vector \( P \) in terms of the resultant \( R \) and angle \( \alpha \) is:
\[ P = R \cos(\alpha) \]
- Substituting the values:
\[ P = 15 \cos(30^\circ) \]
- We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ P = 15 \times \frac{\sqrt{3}}{2} = 15 \times 0.866 = 12.99 \]

3. **Use the Cosine Rule to Find Vector \( Q \):**
- The formula for the component of vector \( Q \) in terms of the resultant \( R \) and angle \( \beta \) is:
\[ Q = R \cos(\beta) \]
- Substituting the values:
\[ Q = 15 \cos(45^\circ) \]
- We know that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ Q = 15 \times \frac{1}{\sqrt{2}} = 15 \times 0.707 = 10.61 \]

4. **Final Result:**
- The magnitudes of the two vectors are:
\[ P \approx 12.99 \quad \text{and} \quad Q \approx 10.61 \]

### Summary:
The two vectors are approximately:
- \( P \approx 12.99 \)
- \( Q \approx 10.61 \)