Two vectors have magnitudes `6` and `8` units, respectively. Find the magnitude of the resultant vector if the angle between vectors is (a) `60^(@)` (b) `90^(@)` and ( c ) `120^(@)`.

Explanation

To find the magnitude of the resultant vector when given two vectors with magnitudes \( a = 6 \) units and \( b = 8 \) units, and the angle \( \theta \) between them, we can use the formula for the magnitude of the resultant vector:

\[ R = \sqrt{a^2 + b^2 + 2ab \cos \theta} \]

We will calculate the resultant for the three angles: \( 60^\circ \), \( 90^\circ \), and \( 120^\circ \).

### Part (a): Angle \( 60^\circ \)

1. **Substitute the values into the formula:**
\[ R = \sqrt{6^2 + 8^2 + 2 \cdot 6 \cdot 8 \cdot \cos 60^\circ} \]

2. **Calculate \( \cos 60^\circ \):**
\[ \cos 60^\circ = \frac{1}{2} \]

3. **Calculate \( R \):**
\[ R = \sqrt{36 + 64 + 2 \cdot 6 \cdot 8 \cdot \frac{1}{2}} \]
\[ R = \sqrt{36 + 64 + 48} \]
\[ R = \sqrt{148} \]
\[ R \approx 12.17 \text{ units} \]

### Part (b): Angle \( 90^\circ \)

1. **Substitute the values into the formula:**
\[ R = \sqrt{6^2 + 8^2 + 2 \cdot 6 \cdot 8 \cdot \cos 90^\circ} \]

2. **Calculate \( \cos 90^\circ \):**
\[ \cos 90^\circ = 0 \]

3. **Calculate \( R \):**
\[ R = \sqrt{36 + 64 + 2 \cdot 6 \cdot 8 \cdot 0} \]
\[ R = \sqrt{36 + 64} \]
\[ R = \sqrt{100} \]
\[ R = 10 \text{ units} \]

### Part (c): Angle \( 120^\circ \)

1. **Substitute the values into the formula:**
\[ R = \sqrt{6^2 + 8^2 + 2 \cdot 6 \cdot 8 \cdot \cos 120^\circ} \]

2. **Calculate \( \cos 120^\circ \):**
\[ \cos 120^\circ = -\frac{1}{2} \]

3. **Calculate \( R \):**
\[ R = \sqrt{36 + 64 + 2 \cdot 6 \cdot 8 \cdot \left(-\frac{1}{2}\right)} \]
\[ R = \sqrt{36 + 64 - 48} \]
\[ R = \sqrt{52} \]
\[ R \approx 7.21 \text{ units} \]

### Summary of Results:
- For \( 60^\circ \): \( R \approx 12.17 \) units
- For \( 90^\circ \): \( R = 10 \) units
- For \( 120^\circ \): \( R \approx 7.21 \) units