The magnitude of resultant of two vectors acting at an angle of `60^(@)` to each other is `sqrt(21)` units. The first vector has magnitude of 4 units. The magnitude of second vector is _________units.

Explanation

To find the magnitude of the second vector when given the magnitude of the first vector and the resultant vector, we can use the formula for the magnitude of the resultant of two vectors acting at an angle to each other.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnitude of the first vector \( A = 4 \) units
- Magnitude of the resultant vector \( R = \sqrt{21} \) units
- Angle between the two vectors \( \theta = 60^\circ \)

2. **Use the Formula for Resultant of Two Vectors:**
The formula for the magnitude of the resultant \( R \) of two vectors \( A \) and \( B \) acting at an angle \( \theta \) is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]

3. **Substitute the Known Values into the Formula:**
\[ \sqrt{21} = \sqrt{4^2 + B^2 + 2 \cdot 4 \cdot B \cdot \cos(60^\circ)} \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we can simplify:
\[ \sqrt{21} = \sqrt{16 + B^2 + 4B} \]

4. **Square Both Sides to Eliminate the Square Root:**
\[ 21 = 16 + B^2 + 4B \]

5. **Rearrange the Equation:**
\[ B^2 + 4B + 16 - 21 = 0 \]
\[ B^2 + 4B - 5 = 0 \]

6. **Solve the Quadratic Equation:**
We can use the quadratic formula \( B = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 4, c = -5 \).
\[ B = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \]
\[ B = \frac{-4 \pm \sqrt{16 + 20}}{2} \]
\[ B = \frac{-4 \pm \sqrt{36}}{2} \]
\[ B = \frac{-4 \pm 6}{2} \]

7. **Calculate the Possible Values for B:**
- \( B = \frac{2}{2} = 1 \)
- \( B = \frac{-10}{2} = -5 \) (not valid since magnitude cannot be negative)

8. **Conclusion:**
The magnitude of the second vector \( B \) is \( 1 \) unit.

### Final Answer:
The magnitude of the second vector is **1 unit**.