Finding the Acute Angle Between Lines

The problem asks us to determine the acute angle between two given lines. To solve this, we need to find the slopes of the lines and then use the formula for the angle between two lines in coordinate geometry.

Step 1: Find the Slopes of the Lines

The general form of a linear equation is \(Ax + By + C = 0\). This can be rearranged into the slope-intercept form, \(y = mx + c\), where \(m\) is the slope of the line.

Line 1: \({\rm{y}} - \sqrt 3 {\rm{x}} - 5 = 0\)

Rearrange the equation to isolate \(y\):

\({\rm{y}} = \sqrt 3 {\rm{x}} + 5\)

Comparing this with \(y = mx + c\), the slope of the first line, \(m_1\), is \(\sqrt 3\).

\(m_1 = \sqrt 3\)

Line 2: \(\sqrt 3 {\rm{y}} - {\rm{x}} + 6 = 0\)

Rearrange the equation to isolate \(y\):

\(\sqrt 3 {\rm{y}} = {\rm{x}} - 6\)

\({\rm{y}} = \frac{1}{\sqrt 3} {\rm{x}} - \frac{6}{\sqrt 3}\)

Comparing this with \(y = mx + c\), the slope of the second line, \(m_2\), is \(\frac{1}{\sqrt 3}\).

\(m_2 = \frac{1}{\sqrt 3}\)

Step 2: Use the Formula for the Angle Between Two Lines

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by the formula:

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

Substitute the values of \(m_1\) and \(m_2\) we found:

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

Step 3: Calculate the Value of \(\tan \theta\)

First, calculate the numerator and denominator separately.

Numerator: \(\sqrt 3 - \frac{1}{\sqrt 3} = \frac{\sqrt 3 \times \sqrt 3 - 1}{\sqrt 3} = \frac{3 - 1}{\sqrt 3} = \frac{2}{\sqrt 3}\)

Denominator: \(1 + (\sqrt 3) \left(\frac{1}{\sqrt 3}\right) = 1 + 1 = 2\)

Now, substitute these back into the formula for \(\tan \theta\):

\(\tan \theta = \left| \frac{\frac{2}{\sqrt 3}}{2} \right| = \left| \frac{2}{\sqrt 3} \times \frac{1}{2} \right| = \left| \frac{1}{\sqrt 3} \right|\)

Since we are looking for the acute angle, we take the positive value:

\(\tan \theta = \frac{1}{\sqrt 3}\)

Step 4: Find the Angle \(\theta\)

We know that \(\tan 30^\circ = \frac{1}{\sqrt 3}\).

Therefore, the angle \(\theta\) is \(30^\circ\).

\(\theta = 30^\circ\)

This angle is acute (less than \(90^\circ\)), so it is the required angle.

Summary of Steps:

• Identify the equations of the lines.
• Convert equations to slope-intercept form \(y = mx + c\) to find slopes \(m_1\) and \(m_2\).
• Use the angle formula \(\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\).
• Substitute slopes and calculate \(\tan \theta\).
• Find the angle \(\theta\) from the value of \(\tan \theta\).

Revision Table: Key Concepts for Angle Between Lines

Additional Information on Lines and Angles

Understanding the relationship between line equations and their slopes is fundamental in coordinate geometry. The slope of a line \(Ax + By + C = 0\) can also be found directly as \(m = -\frac{A}{B}\) (provided \(B \neq 0\)). For the given equations:

• Line 1: \(-\sqrt 3 {\rm{x}} + {\rm{y}} - 5 = 0\). Here, \(A = -\sqrt 3\), \(B = 1\). Slope \(m_1 = - \frac{-\sqrt 3}{1} = \sqrt 3\).
• Line 2: \(-\rm{x} + \sqrt 3 {\rm{y}} + 6 = 0\). Here, \(A = -1\), \(B = \sqrt 3\). Slope \(m_2 = - \frac{-1}{\sqrt 3} = \frac{1}{\sqrt 3}\).

This confirms the slopes calculated by converting to slope-intercept form.

If \(1 + m_1 m_2 = 0\), it means \(m_1 m_2 = -1\). This condition indicates that the two lines are perpendicular, and the angle between them is \(90^\circ\).

If \(m_1 = m_2\), the lines are parallel, and the angle between them is \(0^\circ\) (or \(180^\circ\)).

The question specifically asked for the acute angle, which is why we used the absolute value in the formula and selected the \(30^\circ\) result instead of the potential obtuse angle (\(180^\circ - 30^\circ = 150^\circ\)).