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Let's solve the two questions based on the image:

### (a) Proof for the Angle Between Two Lines:

Given the slopes \( m_1 \) and \( m_2 \) of two lines, we need to show that the tangent of the angle \( \theta \) between them is:

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

#### Derivation:

1. The formula for the tangent of the angle between two lines with slopes \( m_1 \) and \( m_2 \) is derived from the following trigonometric relation:

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

2. This relation is based on the geometric interpretation of the angle between two intersecting lines. If the lines are represented as \( y = m_1x + c_1 \) and \( y = m_2x + c_2 \), then the tangent of the angle between them comes from the formula:

\[
\tan \theta = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}
\]

where \( \alpha \) and \( \beta \) are the angles that the lines make with the x-axis, and \( \tan \alpha = m_1 \), \( \tan \beta = m_2 \).

3. Finally, the absolute value is introduced because the angle \( \theta \) between the two lines is always taken as positive, and the sign of the slope difference does not matter in this case.

Thus, we have proved:

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

### (b) Finding the Angle Between the Lines:

We are given two lines:

1. \( 3y + 2x = 6 \)
2. \( x - y = 5 \)

#### Step 1: Convert to slope-intercept form \( y = mx + c \)

- For the first equation \( 3y + 2x = 6 \):

\[
3y = -2x + 6 \quad \Rightarrow \quad y = -\frac{2}{3}x + 2
\]

So, the slope \( m_1 = -\frac{2}{3} \).

- For the second equation \( x - y = 5 \):

\[
y = x - 5
\]

So, the slope \( m_2 = 1 \).

#### Step 2: Apply the angle formula

Using the formula for the angle between two lines:

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

Substituting \( m_1 = -\frac{2}{3} \) and \( m_2 = 1 \):

\[
\tan \theta = \left| \frac{-\frac{2}{3} - 1}{1 + \left(-\frac{2}{3}\right)(1)} \right| = \left| \frac{-\frac{5}{3}}{1 - \frac{2}{3}} \right| = \left| \frac{-\frac{5}{3}}{\frac{1}{3}} \right| = 5
\]

So,

\[
\tan \theta = 5
\]

Finally, the angle \( \theta \) is:

\[
\theta = \tan^{-1}(5)
\]

This can be approximated using a calculator:

\[
\theta \approx 78.69^\circ
\]

Thus, the angle between the two lines is approximately \( 78.69^\circ \).

---

Do you need further clarifications or details? 

Here are 5 related questions for further practice:
1. Derive the formula for the angle between two lines from first principles.
2. If two lines are perpendicular, what would the product of their slopes be?
3. How do we find the angle between a line and a coordinate axis?
4. What happens to the angle if both slopes are negative?
5. Can two parallel lines have an angle between them? Why or why not?

**Tip:** When finding the angle between two lines, if the tangent is negative, use the absolute value to ensure you're getting the positive acute angle.

If m1 and m2 are the gradients of two lines respectively and the angle between the two lines is θ. Show that tan θ = ± |(m1 - m2) / (1 + m1*m2)|. Find the angle between the lines 3y + 2x = 6 and x - y = 5.

Explore how to calculate the angle between two lines using their gradients with the formula tan θ = ± |(m1 - m2) / (1 + m1*m2)|. This guide includes a detailed explanation and an example solving for the angle between the lines 3y + 2x = 6 and x - y = 5.

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