Exercise 4.3 1. Determine the gradients of the lines with the fo $$ \begin{array}{lll} ext { leaving your } & ext { b) } 30^{\circ}\end{array} $$ c) $$ 150^{\circ} $$ d) $$ 0^{\circ} $$ a) $$ 45^{\circ} $$

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Exercise 4.3 1. Determine the gradients of the lines with the fo $$ \begin{array}{lll}	ext { leaving your } & 	ext { b) } 30^{\circ}\end{array} $$ c) $$ 150^{\circ} $$ d) $$ 0^{\circ} $$ a) $$ 45^{\circ} $$

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Certainly! Let's tackle **Exercise 4.3, Part 1**, which involves determining the gradients (slopes) of lines that make specific angles with the positive $$ x $$-axis. The gradient of a line is given by the tangent of the angle ($$ \theta $$) the line makes with the positive $$ x $$-axis:

$$
\text{Gradient (m)} = \tan(\theta)
$$

We'll compute the gradient for each given angle:

---

### a) $$ 45^{\circ} $$

$$
m = \tan(45^{\circ}) = 1
$$

**Explanation:** 
At $$ 45^{\circ} $$, the rise and run are equal, resulting in a slope of 1.

---

### b) $$ 30^{\circ} $$

$$
m = \tan(30^{\circ}) = \frac{\sqrt{3}}{3} \approx 0.577
$$

**Explanation:** 
Using the exact value, $$ \tan(30^{\circ}) $$ simplifies to $$ \frac{\sqrt{3}}{3} $$. Numerically, this is approximately 0.577.

---

### c) $$ 150^{\circ} $$

$$
m = \tan(150^{\circ}) = \tan(180^{\circ} - 30^{\circ}) = -\tan(30^{\circ}) = -\frac{\sqrt{3}}{3} \approx -0.577
$$

**Explanation:**
$$ 150^{\circ} $$ is in the second quadrant where the tangent is negative. Since $$ 150^{\circ} = 180^{\circ} - 30^{\circ} $$, we use the identity $$ \tan(180^{\circ} - \theta) = -\tan(\theta) $$.

---

### d) $$ 0^{\circ} $$

$$
m = \tan(0^{\circ}) = 0
$$

**Explanation:**
A line making a $$ 0^{\circ} $$ angle with the positive $$ x $$-axis is horizontal, resulting in a slope of 0.

---

### Summary of Gradients

| **Angle ($$ \theta $$)** | **Gradient (m)**       |
|--------------------------|------------------------|
| $$ 45^{\circ} $$         | $$ 1 $$                |
| $$ 30^{\circ} $$         | $$ \frac{\sqrt{3}}{3} $$ (≈ 0.577) |
| $$ 150^{\circ} $$        | $$ -\frac{\sqrt{3}}{3} $$ (≈ -0.577) |
| $$ 0^{\circ} $$          | $$ 0 $$                |

---

**Note:** These calculations assume standard position angles measured from the positive $$ x $$-axis. The tangent function ($$ \tan $$) provides the slope corresponding to each angle.
The gradients of the lines are as follows: - $$ 45^{\circ} $$: 1 - $$ 30^{\circ} $$: $$ \frac{\sqrt{3}}{3} $$ (approximately 0.577) - $$ 150^{\circ} $$: $$ -\frac{\sqrt{3}}{3} $$ (approximately -0.577) - $$ 0^{\circ} $$: 0

Exercise 4.3 1. Determine the gradients of the lines with the fo \( \begin{array}{lll}\text { leaving your } & \text { b) } 30^{\circ}\end{array} \) c) \( 150^{\circ} \) d) \( 0^{\circ} \) a) \( 45^{\circ} \)

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