If `'alpha'` is the angle made by the line with positive direction of `x`-axis , then slope of a line is defined by `m=tanalpha`. Based upon this , answer the following :
`(i)` What acute angle does the line with slope `(1)/(sqrt(3))` make with the vertical line ?
`(ii)` what is the inclination of a line whose slope is `1` ?

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (i): Finding the acute angle made by the line with slope \( \frac{1}{\sqrt{3}} \) with the vertical line.

1. **Understanding the slope**: The slope \( m \) of a line is given by the formula \( m = \tan(\alpha) \), where \( \alpha \) is the angle made with the positive direction of the x-axis.

2. **Given slope**: We have \( m = \frac{1}{\sqrt{3}} \).

3. **Finding \( \alpha \)**: We need to find \( \alpha \) such that:
\[ \tan(\alpha) = \frac{1}{\sqrt{3}} \]
From trigonometric values, we know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Therefore, \( \alpha = 30^\circ \).

4. **Finding the angle with the vertical**: The angle made with the vertical line is given by:
\[ 90^\circ - \alpha \]
Substituting \( \alpha \):
\[ 90^\circ - 30^\circ = 60^\circ \]

Thus, the acute angle made by the line with slope \( \frac{1}{\sqrt{3}} \) with the vertical line is **60 degrees**.

### Part (ii): Finding the inclination of a line whose slope is \( 1 \).

1. **Given slope**: We have \( m = 1 \).

2. **Finding \( \alpha \)**: We need to find \( \alpha \) such that:
\[ \tan(\alpha) = 1 \]
From trigonometric values, we know that:
\[ \tan(45^\circ) = 1 \]
Therefore, \( \alpha = 45^\circ \).

Thus, the inclination of the line whose slope is \( 1 \) is **45 degrees**.

### Summary of Answers:
- (i) The acute angle made by the line with slope \( \frac{1}{\sqrt{3}} \) with the vertical line is **60 degrees**.
- (ii) The inclination of a line whose slope is \( 1 \) is **45 degrees**.