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 \( Q \) that result in a resultant vector \( R \) with a magnitude of 15 units. The angles between the resultant and the vectors are given as \( 30^\circ \) and \( 45^\circ \) respectively.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( R \) be the resultant vector.
- Let \( P \) be the vector making an angle of \( 30^\circ \) with \( R \).
- Let \( Q \) be the vector making an angle of \( 45^\circ \) with \( R \).
- The angle between \( P \) and \( Q \) can be calculated as \( 180^\circ - (30^\circ + 45^\circ) = 105^\circ \).

2. **Applying the Law of Sines**:
- According to the law of sines in triangle \( PQR \):
\[ \frac{P}{\sin(45^\circ)} = \frac{Q}{\sin(30^\circ)} = \frac{R}{\sin(105^\circ)} \]
- We know \( R = 15 \).

3. **Calculating \( R \) using Law of Sines**:
- First, we need to find \( \sin(105^\circ) \):
\[ \sin(105^\circ) = \sin(90^\circ + 15^\circ) = \cos(15^\circ) \approx 0.9659 \]
- Thus, we can write:
\[ \frac{15}{\sin(105^\circ)} = \frac{P}{\sin(45^\circ)} = \frac{Q}{\sin(30^\circ)} \]

4. **Finding \( P \)**:
- Using \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ P = 15 \cdot \frac{\sin(45^\circ)}{\sin(105^\circ)} = 15 \cdot \frac{\frac{1}{\sqrt{2}}}{0.9659} \approx 15 \cdot 0.7321 \approx 10.98 \]

5. **Finding \( Q \)**:
- Using \( \sin(30^\circ) = \frac{1}{2} \):
\[ Q = 15 \cdot \frac{\sin(30^\circ)}{\sin(105^\circ)} = 15 \cdot \frac{\frac{1}{2}}{0.9659} \approx 15 \cdot 0.5176 \approx 7.76 \]

### Final Result:
- The magnitudes of the two vectors are:
- \( P \approx 10.98 \)
- \( Q \approx 7.76 \)