`DeltaLMN` is right angled at M . If `mangleN=60^(@)`, then tan L = _____.

Options

A. `(1)/(2)`
B. `(1)/(sqrt(3))`
C. `(1)/(sqrt(2))`
D. 2

Answer

To solve the problem, we need to find the value of \(\tan L\) in triangle \( \Delta LMN \), where the triangle is right-angled at \( M \) and \( \angle N = 60^\circ \).

### Step-by-step Solution:

1. **Identify the Angles in Triangle**:
Since triangle \( LMN \) is right-angled at \( M \), we have:
\[ \angle L + \angle M + \angle N = 180^\circ \]
Given that \( \angle M = 90^\circ \) and \( \angle N = 60^\circ \), we can substitute these values into the equation:
\[ \angle L + 90^\circ + 60^\circ = 180^\circ \]

2. **Calculate Angle L**:
Rearranging the equation gives:
\[ \angle L = 180^\circ - 90^\circ - 60^\circ \]
Simplifying this, we find:
\[ \angle L = 30^\circ \]

3. **Find the Value of \(\tan L\)**:
Now that we have \( \angle L = 30^\circ \), we can find \( \tan L \):
\[ \tan L = \tan(30^\circ) \]
From trigonometric values, we know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

4. **Final Answer**:
Therefore, the value of \( \tan L \) is:
\[ \tan L = \frac{1}{\sqrt{3}} \]

### Summary:
The final answer is:
\[ \tan L = \frac{1}{\sqrt{3}} \]