Find the inclination of the lines whose slopes are as follows :
`(i) sqrt(3)` `(ii) 1` (iii) `-(1)/(sqrt(3))`

Answer

To find the inclination of the lines whose slopes are given, we can use the relationship between the slope \( m \) of a line and its angle of inclination \( \theta \) with respect to the positive x-axis. The relationship is given by:

\[ m = \tan(\theta) \]

Now, let's solve the problem step by step for each slope.

### (i) For \( m = \sqrt{3} \)

1. **Set up the equation**:
\[ \tan(\theta) = \sqrt{3} \]

2. **Identify the angle**:
We know that:
\[ \tan(60^\circ) = \sqrt{3} \]
Therefore, we can conclude that:
\[ \theta = 60^\circ \]

### (ii) For \( m = 1 \)

1. **Set up the equation**:
\[ \tan(\theta) = 1 \]

2. **Identify the angle**:
We know that:
\[ \tan(45^\circ) = 1 \]
Therefore, we can conclude that:
\[ \theta = 45^\circ \]

### (iii) For \( m = -\frac{1}{\sqrt{3}} \)

1. **Set up the equation**:
\[ \tan(\theta) = -\frac{1}{\sqrt{3}} \]

2. **Identify the angle**:
We know that:
\[ \tan(150^\circ) = -\frac{1}{\sqrt{3}} \]
This is because:
\[ \tan(150^\circ) = \tan(180^\circ - 30^\circ) = -\tan(30^\circ) = -\frac{1}{\sqrt{3}} \]
Therefore, we can conclude that:
\[ \theta = 150^\circ \]

### Summary of Results:
- For \( m = \sqrt{3} \), \( \theta = 60^\circ \)
- For \( m = 1 \), \( \theta = 45^\circ \)
- For \( m = -\frac{1}{\sqrt{3}} \), \( \theta = 150^\circ \)