What is the inclination of the line `sqrt(3)x - y - 1 = 0` ?

Options

A. `30^(@)`
B. `60^(@)`
C. `135^(@)`
D. `150^(@)`

Answer

To find the inclination of the line given by the equation \( \sqrt{3}x - y - 1 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation into slope-intercept form
The standard form of a line is \( y = mx + c \), where \( m \) is the slope of the line. We need to rearrange the given equation into this form.

Starting with the equation:
\[ \sqrt{3}x - y - 1 = 0 \]

Add \( y \) and \( 1 \) to both sides:
\[ \sqrt{3}x - 1 = y \]

Now, we can write it as:
\[ y = \sqrt{3}x - 1 \]

### Step 2: Identify the slope
From the equation \( y = \sqrt{3}x - 1 \), we can see that the slope \( m \) is \( \sqrt{3} \).

### Step 3: Relate the slope to the angle of inclination
The slope \( m \) can be related to the angle of inclination \( \theta \) using the formula:
\[ m = \tan(\theta) \]

Substituting the value of \( m \):
\[ \sqrt{3} = \tan(\theta) \]

### Step 4: Find the angle \( \theta \)
To find \( \theta \), we take the inverse tangent (arctan) of \( \sqrt{3} \):
\[ \theta = \tan^{-1}(\sqrt{3}) \]

From trigonometric values, we know:
\[ \tan(60^\circ) = \sqrt{3} \]

Thus,
\[ \theta = 60^\circ \]

### Conclusion
The inclination of the line \( \sqrt{3}x - y - 1 = 0 \) is \( 60^\circ \).

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