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8.07 Extension: Relating the tangent to the slope 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 $xy$xy-plane.  

If we wanted to find the slope, 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 slope 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 slope.

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

Connection between slope and tangent ratio

For a line with slope $m$m and angle of inclination $\theta$θ made with the positive $x$x-axis,



Finding the angle of inclination from the slope

Worked example

Question 1

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

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

We also know that it is quite a bit steeper than the line $y=x$y=x, which has slope $1$1.  We can also find the angle of inclination from this value. Remember that the value of the slope, 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 slope from the angle of inclination

Worked example

Question 2

Evaluate: Suppose we are told that the angle of inclination is $45^{\circ}$45.  Use this information to determine the slope 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 step (2), how by making the adjacent length $1$1 unit we were able to remove the fractional piece more easily?


Practice questions

Question 3

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 slope $m$m of the line.

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

Question 4

A line has slope measuring $-\frac{1}{\sqrt{3}}$13, and makes an angle $\theta$θ, counterclockwise from the positive $x$x-axis to the line.

Solve for $\theta$θ in degrees.

Enter each line as an equation.

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