Find the equations to the straight lines which pass through the origin and are inclined at `75^(@)` to the straight line
`x+ y + sqrt""3 (y-x)=a`.

Answer

To find the equations of the straight lines that pass through the origin and are inclined at \(75^\circ\) to the given line \(x + y + \sqrt{3}(y - x) = a\), we will follow these steps:

### Step 1: Rearranging the Given Line
First, we need to rearrange the given equation to find its slope. The equation is:
\[ x + y + \sqrt{3}(y - x) = a \]
Expanding this, we have:
\[ x + y + \sqrt{3}y - \sqrt{3}x = a \]
Combining like terms:
\[ (1 - \sqrt{3})x + (1 + \sqrt{3})y = a \]

### Step 2: Finding the Slope of the Given Line
From the rearranged equation, we can express it in the slope-intercept form \(y = mx + c\):
\[ (1 + \sqrt{3})y = a + (1 - \sqrt{3})x \]
Thus, the slope \(m_1\) of the given line is:
\[ m_1 = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]

### Step 3: Using the Angle Between Lines Formula
Let \(m_2\) be the slope of the line we want to find. The angle \(\theta\) between the two lines is given as \(75^\circ\). The formula for the tangent of the angle between two lines is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \(\theta = 75^\circ\):
\[ \tan 75^\circ = 2 + \sqrt{3} \]
Thus, we have:
\[ 2 + \sqrt{3} = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 4: Setting Up the Equation
Substituting \(m_1\) into the equation:
\[ 2 + \sqrt{3} = \left| \frac{\frac{1 - \sqrt{3}}{1 + \sqrt{3}} - m_2}{1 + \frac{1 - \sqrt{3}}{1 + \sqrt{3}} m_2} \right| \]

### Step 5: Solving for \(m_2\)
We will solve the equation for \(m_2\) considering both cases of the modulus:
1. \(2 + \sqrt{3} = \frac{m_1 - m_2}{1 + m_1 m_2}\)
2. \(2 + \sqrt{3} = -\frac{m_1 - m_2}{1 + m_1 m_2}\)

### Step 6: Finding the Two Possible Slopes
After solving these equations, we will find two possible values for \(m_2\):
1. \(m_2 = -\sqrt{3}\)
2. \(m_2 = \infty\) (which corresponds to a vertical line)

### Step 7: Writing the Equations of the Lines
1. For \(m_2 = -\sqrt{3}\):
\[ y = -\sqrt{3}x \implies \sqrt{3}x + y = 0 \]
2. For \(m_2 = \infty\):
\[ x = 0 \quad \text{(the y-axis)} \]

### Final Answer
The equations of the lines that pass through the origin and are inclined at \(75^\circ\) to the given line are:
1. \(\sqrt{3}x + y = 0\)
2. \(x = 0\) (the y-axis)

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