topic badge
New Zealand
Level 7 - NCEA Level 2

Trigonometry and gradients of lines



The slope of a line is a measure of how steep it is and as we know this is called the gradient. 

A gradient is a single value that describes:

  • if a line is increasing (has positive gradient)
  • if a line is decreasing (has negative gradient)
  • how far up or down the line moves (change in the $y$y value) with every unit step to the right (for every 1 unit increase in $x$x)

If you have a graph, like this one, you can find the rise and run by drawing a right triangle created by any two points on the line. The line itself becomes the hypotenuse. 


In this case, the gradient is positive because, over the $3$3 unit increase in the $x$x values, the $y$y value has increased.


Relating the tangent to the gradient of a line

A special relationship exists between the tangent ratio and the slope of a line.


Pictured here is a straight line on the Cartesian plane.  

If we wanted to find the gradient, we could look at $\frac{\text{rise }}{\text{run }}$rise run , which in this line is $\frac{AB}{CB}$ABCB.

Look at angle $\theta$θ, what is $\tan\theta$tanθ?

$\tan\theta=\frac{AB}{CB}$tanθ=ABCB.  This is the value of the gradient we figured out just before!

This means that $\tan\theta$tanθ, or specifically the tangent of the angle the line makes with the $x$x-axis in the positive direction, is equal to the gradient.

The angle $\theta$θ is called the angle of inclination.


Finding the angle of inclination from the gradient

Worked example

Evaluate: Suppose we know that a straight line has a gradient of $3$3.  What does this tell us about the line?

Think: Well we know that it is increasing, (positive gradient), so has this shape.

We also know that it is quite a bit steeper than the line $y=x$y=x, which has gradient $1$1.  We can also find the angle of inclination from this value. Remember that the value of the gradient, in this case $3$3, is a measure of the $\frac{rise}{run}$riserun, so in this case the rise is $3$3, and the run is $1$1 as $3=\frac{3}{1}$3=31.

Do: We can draw a diagram of a triangle like this:

see how the rise and run form the vertical and horizontal components of our triangle.  From this, we can calculate the angle of inclination, which is the angle marked $\theta$θ on this diagram.

$\tan\left(\theta\right)$tan(θ) $=$= $\frac{opposite}{adjacent}$oppositeadjacent
$\tan\left(\theta\right)$tan(θ) $=$= $\frac{3}{1}$31
$\theta$θ $=$= $\tan^{-1}\left(3\right)$tan1(3)
$\theta$θ $=$= $71.565^{\circ}$71.565

Reflect: So the angle of inclination is $71.565^{\circ}$71.565


Finding the gradient from the angle of inclination

Worked example

Evaluate: Suppose we are told that the angle of inclination is $45^{\circ}$45.  Use this information to determine the gradient of the line.

Think: In this case, we start with the right triangle, the angle is given ($45^{\circ}$45) and we will add the adjacent side length of $1$1 unit.  This is the horizontal run.

Do: Finding the vertical distance (the opposite side) is a process of trigonometry.

$\tan\theta$tanθ $=$= $\frac{opposite}{adjacent}$oppositeadjacent  
$\tan\left(45^{\circ}\right)$tan(45) $=$= $\frac{opposite}{1}$opposite1 (2)
$\tan\left(45^{\circ}\right)$tan(45) $=$= $opposite$opposite  
$opposite$opposite $=$= $1$1  

Reflect: See on row 2, how by making the adjacent length 1 unit we were able to remove the fractional piece more easily?

Worked Examples


Find the gradient of a line that is inclined at an angle of 50° to the positive $x$x-axis . Write your answer correct to 1 decimal place.


The given line has an angle of inclination of $\theta$θ with the positive $x$x-axis.

  1. Determine the gradient of the line.

  2. Hence solve for $\theta$θ. Give your answer in degrees to two decimal places.


Two points $A$A$\left(1,-2\right)$(1,2) and $B$B$\left(9,30\right)$(9,30) lie on a line that makes an angle of $\theta$θ degrees with the positive $x$x-axis.

  1. Determine the gradient $m$m of the line.

  2. Find $\theta$θ in degrees to two decimal places.




Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions


Apply trigonometric relationships in solving problems

What is Mathspace

About Mathspace