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 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.
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.
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 | $=$= | $\tan^{-1}\left(3\right)$tan−1(3) |
$\theta$θ | $=$= | $71.565^{\circ}$71.565∘ |
Reflect: So the angle of inclination is $71.565^{\circ}$71.565∘
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?
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.
Determine the gradient of the line.
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.
Determine the gradient $m$m of the line.
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