Two tangents making an angle of `60^@` between them, are drawn to a circle of radius `sqrt2` cm, then find the length of each tangent.

Answer

To find the length of each tangent drawn to a circle of radius \(\sqrt{2}\) cm, making an angle of \(60^\circ\) between them, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let \(C\) be the center of the circle, and \(A\) be the point where the tangents touch the circle. The points where the tangents touch the circle can be denoted as \(B\) and \(D\). The angle between the two tangents \(AB\) and \(AD\) is given as \(60^\circ\).

2. **Bisect the Angle**:
- Since the tangents \(AB\) and \(AD\) meet at point \(A\) and form an angle of \(60^\circ\), the angle between the radius \(CB\) (or \(CD\)) and the tangent \(AB\) (or \(AD\)) is \(90^\circ\). Therefore, the angle \(CAB\) is half of \(60^\circ\), which is \(30^\circ\).

3. **Use Right Triangle Properties**:
- In triangle \(CBA\), we have:
- \(CB = \text{radius} = \sqrt{2}\) cm
- \(\angle CAB = 30^\circ\)

4. **Apply the Tangent Function**:
- Using the tangent function in triangle \(CBA\):
\[ \tan(30^\circ) = \frac{BC}{AB} \]
- We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).

5. **Set Up the Equation**:
- Substitute the known values into the equation:
\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{2}}{AB} \]

6. **Solve for \(AB\)**:
- Rearranging the equation gives:
\[ AB = \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \text{ cm} \]

7. **Conclusion**:
- Therefore, the length of each tangent \(AB\) and \(AD\) is \(\sqrt{6}\) cm.

### Final Answer:
The length of each tangent is \(\sqrt{6}\) cm.