The equation of the straight line which makes angle of `15^(@)` with the positive direction of x-axis and which cuts an intercept of length 4 on the negative direction of y-axis, is

Options

A. `y=(2-sqrt3)x-4`
B. `y=(2+sqrt3)x+4`
C. `y=(2-sqrt3)x+4`
D. `y=(2+sqrt3)x-4`

Answer

To find the equation of the straight line that makes an angle of \(15^\circ\) with the positive direction of the x-axis and cuts an intercept of length 4 on the negative direction of the y-axis, we can follow these steps:

### Step 1: Determine the slope of the line
The slope \(m\) of a line that makes an angle \(\theta\) with the positive x-axis is given by:
\[ m = \tan(\theta) \]
For \(\theta = 15^\circ\):
\[ m = \tan(15^\circ) \]

### Step 2: Use the tangent subtraction formula
We can express \(\tan(15^\circ)\) using the tangent subtraction formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} \]
Where \(\tan(45^\circ) = 1\) and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Step 3: Rationalize the expression
To simplify \(\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\), we multiply the numerator and denominator by \(\sqrt{3} - 1\):
\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \]

### Step 4: Write the equation of the line
The line cuts the y-axis at \(-4\), so the y-intercept \(b = -4\). The equation of the line in slope-intercept form is:
\[ y = mx + b \]
Substituting \(m\) and \(b\):
\[ y = (2 - \sqrt{3})x - 4 \]

### Step 5: Rearranging the equation
We can rearrange the equation to standard form:
\[ y + 4 = (2 - \sqrt{3})x \]

### Final Equation
Thus, the equation of the line is:
\[ y = (2 - \sqrt{3})x - 4 \]