Find the equation of a line whose inclination with the x-axis is `30^(@)` and which passes through the point (0, 5).

Answer

To find the equation of a line whose inclination with the x-axis is \(30^\circ\) and which passes through the point \((0, 5)\), we can follow these steps:

### Step 1: Determine the slope of the line
The slope \(m\) of a line can be found using the tangent of the angle of inclination. The formula is:
\[ m = \tan(\theta) \]
For \(\theta = 30^\circ\):
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 2: Use the point-slope form of the equation of a line
The point-slope form of the equation of a line is given by:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is a point on the line. Here, we have the point \((0, 5)\) and the slope \(m = \frac{1}{\sqrt{3}}\). Substituting these values into the equation gives:
\[ y - 5 = \frac{1}{\sqrt{3}}(x - 0) \]
This simplifies to:
\[ y - 5 = \frac{1}{\sqrt{3}}x \]

### Step 3: Rearranging the equation
To express the equation in standard form, we can rearrange it:
\[ y = \frac{1}{\sqrt{3}}x + 5 \]
Now, we can multiply through by \(\sqrt{3}\) to eliminate the fraction:
\[ \sqrt{3}y = x + 5\sqrt{3} \]
Rearranging gives:
\[ x - \sqrt{3}y + 5\sqrt{3} = 0 \]

### Final Equation
Thus, the equation of the line is:
\[ x - \sqrt{3}y + 5\sqrt{3} = 0 \]