Calculating and Comparing Angles Between Lines

The question asks us to find the angle between two pairs of lines and then compare these angles. The first angle is denoted by $\alpha$ and the second (acute) angle is denoted by $\beta$. We will calculate each angle separately using the slopes of the lines.

Finding the Angle $\alpha$

The first pair of lines is:

1. Line 1: $x + y - 3 = 0$
2. Line 2: $x - y + 3 = 0$

To find the angle between these lines, we first determine their slopes. The general form of a linear equation is $Ax + By + C = 0$, and its slope is given by $-A/B$. Alternatively, we can rewrite the equations in the slope-intercept form, $y = mx + c$, where $m$ is the slope.

• For Line 1 ($x + y - 3 = 0$):
  Rewrite as $y = -x + 3$. The slope is $m_1 = -1$.
• For Line 2 ($x - y + 3 = 0$):
  Rewrite as $y = x + 3$. The slope is $m_2 = 1$.

Now we use the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substitute the slopes $m_1 = -1$ and $m_2 = 1$ to find $\alpha$:

$\tan \alpha = \left| \frac{-1 - 1}{1 + (-1)(1)} \right| = \left| \frac{-2}{1 - 1} \right| = \left| \frac{-2}{0} \right|$

The denominator is zero, which means the lines are perpendicular. The angle between perpendicular lines is $90^\circ$ or $\frac{\pi}{2}$ radians.

So, $\alpha = 90^\circ$.

Finding the Acute Angle $\beta$

The second pair of lines is:

1. Line 3: $x - \sqrt{3} y + 2 \sqrt{3} = 0$
2. Line 4: $\sqrt{3} x - y + 1 = 0$

Again, let's find their slopes:

• For Line 3 ($x - \sqrt{3} y + 2 \sqrt{3} = 0$):
  Rewrite as $\sqrt{3} y = x + 2 \sqrt{3}$. Divide by $\sqrt{3}$: $y = \frac{1}{\sqrt{3}} x + 2$. The slope is $m_3 = \frac{1}{\sqrt{3}}$.
• For Line 4 ($\sqrt{3} x - y + 1 = 0$):
  Rewrite as $y = \sqrt{3} x + 1$. The slope is $m_4 = \sqrt{3}$.

Now use the angle formula with $m_3 = \frac{1}{\sqrt{3}}$ and $m_4 = \sqrt{3}$ to find $\beta$ (acute angle):

$\tan \beta = \left| \frac{m_3 - m_4}{1 + m_3 m_4} \right|$

Substitute the slopes:

$\tan \beta = \left| \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{1 + \left(\frac{1}{\sqrt{3}}\right) (\sqrt{3})} \right|$

Calculate the numerator and denominator:

• Numerator: $\frac{1}{\sqrt{3}} - \sqrt{3} = \frac{1 - \sqrt{3} \times \sqrt{3}}{\sqrt{3}} = \frac{1 - 3}{\sqrt{3}} = \frac{-2}{\sqrt{3}}$
• Denominator: $1 + \left(\frac{1}{\sqrt{3}}\right) (\sqrt{3}) = 1 + 1 = 2$

Now substitute these back into the formula for $\tan \beta$:

$\tan \beta = \left| \frac{\frac{-2}{\sqrt{3}}}{2} \right| = \left| \frac{-2}{2\sqrt{3}} \right| = \left| \frac{-1}{\sqrt{3}} \right| = \frac{1}{\sqrt{3}}$

Since $\tan \beta = \frac{1}{\sqrt{3}}$ and $\beta$ is the acute angle, we know that $\beta = 30^\circ$ or $\frac{\pi}{6}$ radians (because $\tan 30^\circ = \frac{1}{\sqrt{3}}$).

So, $\beta = 30^\circ$.

Comparing $\alpha$ and $\beta$

We found that:

• $\alpha = 90^\circ$
• $\beta = 30^\circ$

Comparing these two values, we see that $90^\circ > 30^\circ$.

Therefore, $\alpha > \beta$.

Revision Table: Angle Between Lines

Additional Information: Angles and Slopes

The angle between two lines is a fundamental concept in coordinate geometry. The slope of a line is directly related to the angle the line makes with the positive x-axis. If a line makes an angle $\theta'$ with the positive x-axis, its slope $m$ is given by $m = \tan \theta'$. The angle between two lines can then be derived from the difference between the angles they make with the x-axis.

The formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ gives the acute angle between the lines. If the non-acute angle is required, the absolute value is removed, and the angle can be found using $\tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}$. The signs of $m_1, m_2$ and their product $m_1 m_2$ are crucial for determining the specific angle.

In this problem, the acute angle $\beta$ was specifically requested, which is why the absolute value in the formula is important for its calculation.