Gradient and the Angle of Inclination

A line rises $3$ units for every $4$ across. Sketch the right triangle that sits under it, with the horizontal along the x-axis. What is the tangent of the angle at the x-axis? Now compare that with the gradient of the line.

Drop a right triangle under any line, with one leg along the x-axis. Its opposite side is the rise and its adjacent side is the run. So the tangent of the angle at the x-axis is rise over run, which is exactly the gradient.

Every line that is not vertical crosses the x-axis, or is parallel to it. The angle of inclination is the angle you turn through, anticlockwise, from the positive direction of the x-axis until you are pointing along the line.

By convention it is taken in the range $0° \leq \theta < 180°$. A horizontal line has $\theta = 0°$. A line rising to the right has an acute angle. A line falling to the right has an obtuse angle, because you have to turn past the vertical to point along it.

The angle belongs to the line, not to any particular pair of points on it. Slide along the line as far as you like and the direction is unchanged, which is why the gradient is constant too.

Take any two points on the line and complete the right triangle whose horizontal leg lies along the x-direction. The vertical leg is the rise and the horizontal leg is the run.

The angle in that triangle at the left-hand vertex equals the angle of inclination, because the horizontal leg is parallel to the x-axis and the line is a transversal, so they are corresponding angles.

In that triangle the rise is opposite the angle and the run is adjacent to it, so

$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\text{rise}}{\text{run}} = m$

This is not a new formula so much as a translation. Gradient and tangent were always measuring the same thing.

A line falling to the right has a negative gradient, and $\tan\theta$ is negative for obtuse angles, so the relationship still holds. The difficulty is the calculator.

Ask a calculator for $\tan^{-1}(-0.5)$ and it returns $-26.6°$. That is a legitimate angle with the right tangent, but it is outside the agreed range for an angle of inclination.

The fix is one step. Tangent repeats every $180°$, so adding $180°$ gives another angle with the same tangent, and lands inside the range:

$\theta = -26.6° + 180° = 153.4°$

You can check it against the picture. A gradient of $-0.5$ is a gentle fall, so the line points slightly below the horizontal going right, which is a little under $180°$ measured anticlockwise. If your answer for a falling line is not between $90°$ and $180°$, you have not finished.

From equation to angle: read the gradient off the equation, take the inverse tangent, adjust if the answer is negative.

From angle to equation: take the tangent of the angle to get the gradient, then use a point with $y - y_1 = m(x - x_1)$.

The second direction is how trades work. A roof pitched at $22°$ has gradient $\tan 22° = 0.404$, which is a rise of about $404$ mm per metre of run. A road sign reading $10\%$ is quoting the gradient as a percentage, $m = 0.1$, so its angle is $\tan^{-1}(0.1) = 5.7°$.

Note how small that angle is. Road gradients that feel severe are only a few degrees, which is worth remembering when a diagram tempts you into an answer of $45°$.

Parallel lines point the same way, so they have the same angle of inclination and therefore the same gradient. That much is unsurprising.

Perpendicular lines are more interesting. If one has inclination $\theta$, the other has inclination $\theta + 90°$, or $\theta - 90°$ if that would push it past $180°$.

Rotating a point $90°$ anticlockwise about the origin sends $(x, y)$ to $(-y, x)$. Applying this to the unit circle point $(\cos\theta, \sin\theta)$ gives $(-\sin\theta, \cos\theta)$, so $\cos(\theta + 90°) = -\sin\theta$ and $\sin(\theta + 90°) = \cos\theta$. Hence

$\tan(\theta + 90°) = \dfrac{\cos\theta}{-\sin\theta} = -\dfrac{1}{\tan\theta}$

So $m_2 = -\dfrac{1}{m_1}$, which rearranges to the familiar $m_1m_2 = -1$. The perpendicular gradient rule from coordinate geometry is a trigonometric identity in disguise.

Five items translating between gradient and angle of inclination. Predict whether the angle is acute or obtuse before you calculate.

1. 1 A line has gradient $1$. Find its angle of inclination.

   Show answer
   $\tan\theta = 1$, and the tangent of $45°$ is $1$.$45°$

2. 2 A line has gradient $\sqrt{3}$. Find its angle of inclination exactly.

   Show answer
   An exact value from the $30$-$60$-$90$ triangle.$60°$

3. 3 A line has an angle of inclination of $135°$. Find its gradient.

   Show answer
   $\tan 135° = -\tan 45°$ by the supplementary relationship.$m = -1$

4. 4 Find the angle of inclination of the line through $(0, 0)$ and $(5, 3)$, to one decimal place.

   Show answer
   $m = \tfrac{3}{5} = 0.6$, then $\tan^{-1}(0.6)$.$31.0°$

5. 5 A line has gradient $-\dfrac{1}{\sqrt{3}}$. Find its angle of inclination exactly.

   Show answer
   The related acute angle is $30°$, and the gradient is negative, so the angle is obtuse.$150°$