9.10 Gradient of a line using trigonometric ratios

Gradient

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.

Exploration

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

example 1

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	$=$=	$\tan^{-1}\left(3\right)$tan−1(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

example 2

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$2, how by making the adjacent length $1$1 unit we were able to remove the fractional piece more easily?

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3