Angle between y = x + 6 and y = `sqrt(3)x + 7` is

Options

A. `75^(@)`
B. `45^(@)`
C. `15^(@)`
D. `30^(@)`

Answer

To find the angle between the two lines given by the equations \( y = x + 6 \) and \( y = \sqrt{3}x + 7 \), we can follow these steps:

### Step 1: Identify the slopes of the lines
The equations of the lines are in the slope-intercept form \( y = mx + c \), where \( m \) is the slope.

- For the first line \( y = x + 6 \):
- The slope \( m_1 = 1 \).

- For the second line \( y = \sqrt{3}x + 7 \):
- The slope \( m_2 = \sqrt{3} \).

### Step 2: Use the formula for the angle between two lines
The formula to find the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:

\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the values of slopes into the formula
Substituting \( m_1 = 1 \) and \( m_2 = \sqrt{3} \):

\[ \tan(\theta) = \left| \frac{1 - \sqrt{3}}{1 + 1 \cdot \sqrt{3}} \right| \]

### Step 4: Simplify the expression
Now, simplify the numerator and denominator:

- Numerator: \( 1 - \sqrt{3} \)
- Denominator: \( 1 + \sqrt{3} \)

Thus, we have:

\[ \tan(\theta) = \left| \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \right| \]

### Step 5: Determine the angle
Since we are interested in the angle \( \theta \) in the first quadrant, we can drop the absolute value (as both \( 1 - \sqrt{3} \) and \( 1 + \sqrt{3} \) will yield a positive value):

\[ \tan(\theta) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]

Now, we can recognize that:

\[ \tan(15^\circ) = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \]

Thus, we find:

\[ \theta = 15^\circ \]

### Conclusion
The angle between the lines \( y = x + 6 \) and \( y = \sqrt{3}x + 7 \) is \( 15^\circ \).

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