To solve for the gradient of a straight line inclined at a given angle $\theta$ to the positive direction of the $x$-axis, we use the formula:

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

Now let’s calculate the gradient for each option:

a. $\theta = 45^\circ$:
$\text{Gradient} = \tan(45^\circ) = 1$

b. $\theta = 135^\circ$:
$\text{Gradient} = \tan(135^\circ) = -1$
(As $135^\circ$ lies in the second quadrant where tangent is negative.)

c. $\theta = 60^\circ$:
$\text{Gradient} = \tan(60^\circ) = \sqrt{3}$

d. $\theta = 120^\circ$:
$\text{Gradient} = \tan(120^\circ) = -\sqrt{3}$
(As $120^\circ$ lies in the second quadrant where tangent is negative.)

Final Answers:

• a: $1$
• b: $-1$
• c: $\sqrt{3}$
• d: $-\sqrt{3}$

Let me know if you need further explanations or clarification!

Related Questions:

1. How is the tangent function defined for angles in different quadrants?
2. What is the relationship between the angle of inclination and the gradient of a line?
3. How do you determine the quadrant in which an angle lies?
4. How do you calculate the gradient for angles greater than $180^\circ$?
5. What are the properties of gradients for horizontal and vertical lines?

Tip:

The tangent function is positive in the first and third quadrants and negative in the second and fourth quadrants. Always check the quadrant when calculating gradients!