The slope as the tangent of an angle

Geometry, grade 10 (4 units) Lesson 14 of 20Learning

The slope as the tangent of an angle

The slope and the angle

A line not parallel to the axes makes an acute angle $\displaystyle \alpha$ with the $\displaystyle x$ -axis. The change in $\displaystyle x$ and the change in $\displaystyle y$ between two of its points are the legs of a right-angled triangle in which $\displaystyle \alpha$ is one of the angles, so their ratio – the slope – is its tangent:

\displaystyle m

\displaystyle =

\displaystyle \tan\alpha

\displaystyle y=2x-2

\displaystyle A=(1,0)

\displaystyle B=(3,4)

In the drawing, the line $\displaystyle {y}={2x-2}$ crosses the $\displaystyle x$ -axis at $\displaystyle A(1,0)$ :

from $\displaystyle A$ to $\displaystyle B(3,4)$ : the leg along the $\displaystyle x$ -axis is $\displaystyle 2$ , and the leg perpendicular to it $\displaystyle 4$ ;
$\displaystyle {\tan\alpha}={\frac{4}{2}}={2}$ – the slope;
$\displaystyle {\alpha}={\tan^{-1}2\approx63.43^{\circ}}$ .

A falling line

A falling line has a negative slope, and the acute angle between it and the $\displaystyle x$ -axis gives the size of the slope: the slope is $\displaystyle -\tan\alpha$ .

$\displaystyle {y}={x}$ – a rising line, $\displaystyle {\tan\alpha}={1}$ , and $\displaystyle {\alpha}={45^{\circ}}$ ;
$\displaystyle {y}={-\sqrt{3}x+2}$ – a falling line, $\displaystyle {\tan\alpha}={\sqrt{3}}$ , and $\displaystyle {\alpha}={60^{\circ}}$ ;
a line parallel to the $\displaystyle x$ -axis – slope $\displaystyle {0}={\tan0^{\circ}}$ ; a line parallel to the $\displaystyle y$ -axis – an angle of $\displaystyle 90^{\circ}$ , and $\displaystyle \tan90^{\circ}$ is not defined, nor is its slope.

Exercise 1

Find the acute angle between the line and the $\displaystyle x$ -axis.
Exactly where possible; otherwise – rounded to two decimal places.

\displaystyle {y}={x+3}

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\displaystyle {y}={3x}

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\displaystyle {y}={-0.5x+1}

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Exercise 2

A rising line passes through the point $\displaystyle P(1,2)$ , and the angle between it and the $\displaystyle x$ -axis is $\displaystyle 45^{\circ}$ .
Write the line’s equation.

\displaystyle P=(1,2)

Exercise 3

A falling line passes through the origin, and the angle between it and the $\displaystyle x$ -axis is $\displaystyle 60^{\circ}$ .
Write the line’s equation, exactly.

The angles of a triangle in coordinates

When a side of a triangle is on the $\displaystyle x$ -axis, each of the two angles next to it is the angle between another side and the $\displaystyle x$ -axis – and that side’s slope gives it.

Example: $\displaystyle A(-2,0)$ , $\displaystyle B(4,0)$ , $\displaystyle C(0,4)$ .

$\displaystyle {m_{AC}}={\frac{4-0}{0-(-2)}}={2}$ , so $\displaystyle {\angle A}={\tan^{-1}2\approx63.43^{\circ}}$ ;
$\displaystyle {m_{BC}}={\frac{4-0}{0-4}}={-1}$ : a falling line, and $\displaystyle {\tan\angle B}={1}$ , so $\displaystyle {\angle B}={45^{\circ}}$ ;
$\displaystyle {\angle C}={180^{\circ}-63.43^{\circ}-45^{\circ}}={71.57^{\circ}}$ .

Exercise 4

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(0,0)$ , $\displaystyle B(6,0)$ and $\displaystyle C(4,5)$ .
Find the triangle’s angles, rounded to two decimal places.

\displaystyle \angle A

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\displaystyle \angle B

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\displaystyle \angle C

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Exercise 5

The board shows the lines $\displaystyle {y}={x}$ and $\displaystyle {y}={\frac{1}{2}x}$ .
Find the angle between them: the difference between the angles they make with the $\displaystyle x$ -axis.
Is the tangent of that angle equal to the difference between the slopes, $\displaystyle 1-\frac{1}{2}$ ?
What do you conclude?

\displaystyle y=x

\displaystyle y=\frac{1}{2}x

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Formulas: geometry, grade 10 (4 units)

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Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$