Find the equation of the straight line which passes through the origin and makes angle `60^0` with the line `x+sqrt(3)y+sqrt(3)=0` .

Explanation

To find the equation of the straight line that passes through the origin and makes an angle of \(60^\circ\) with the line given by the equation \(x + \sqrt{3}y + \sqrt{3} = 0\), we can follow these steps:

### Step 1: Find the slope of the given line
The equation of the line can be rewritten in the slope-intercept form \(y = mx + c\).

Starting from:
\[ x + \sqrt{3}y + \sqrt{3} = 0 \]
we can rearrange it to find \(y\):
\[ \sqrt{3}y = -x - \sqrt{3} \]
\[ y = -\frac{1}{\sqrt{3}}x - 1 \]

From this, we can see that the slope \(m_1\) of the given line is:
\[ m_1 = -\frac{1}{\sqrt{3}} \]

### Step 2: Use the angle formula to find the slope of the new line
We know that the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Given that \(\theta = 60^\circ\), we have:
\[ \tan(60^\circ) = \sqrt{3} \]

Substituting \(m_1\) and \(\tan(60^\circ)\) into the formula:
\[ \sqrt{3} = \left| \frac{m_2 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}m_2} \right| \]

### Step 3: Solve for \(m_2\)
We can consider two cases for the absolute value.

**Case 1:**
\[ \sqrt{3} = \frac{m_2 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}m_2} \]
Cross-multiplying gives:
\[ \sqrt{3}(1 - \frac{1}{\sqrt{3}}m_2) = m_2 + \frac{1}{\sqrt{3}} \]
Expanding and rearranging:
\[ \sqrt{3} - m_2 = m_2 + \frac{1}{\sqrt{3}} \]
\[ \sqrt{3} - \frac{1}{\sqrt{3}} = 2m_2 \]
\[ m_2 = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{2} = \frac{3 - 1}{2\sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]

**Case 2:**
\[ \sqrt{3} = -\frac{m_2 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}m_2} \]
Cross-multiplying gives:
\[ -\sqrt{3}(1 - \frac{1}{\sqrt{3}}m_2) = m_2 + \frac{1}{\sqrt{3}} \]
Expanding and rearranging:
\[ -\sqrt{3} + m_2 = m_2 + \frac{1}{\sqrt{3}} \]
This leads to a contradiction, indicating that this case does not yield a valid slope.

### Step 4: Form the equations of the lines
We have found one slope \(m_2 = \frac{1}{\sqrt{3}}\). The line passing through the origin with this slope is:
\[ y = \frac{1}{\sqrt{3}}x \]

The second line that makes an angle of \(60^\circ\) with the given line can be found using the negative slope:
\[ m_2 = -\frac{1}{\sqrt{3}} - \sqrt{3} = -\frac{1 + 3}{\sqrt{3}} = -\frac{4}{\sqrt{3}} \]
The line with this slope is:
\[ y = -\frac{4}{\sqrt{3}}x \]

### Final Answer
The two equations of the lines that pass through the origin and make an angle of \(60^\circ\) with the given line are:
1. \(y = \frac{1}{\sqrt{3}}x\)
2. \(y = -\frac{4}{\sqrt{3}}x\)