Right-triangle trigonometry

The word trigonometry derives from the Greek for "triangle measure", and that's a pretty good description of the nuts and bolts of it. Still, that's not all there is to trigonometry, by a long way. Learning trig (we call it "trig") will crack the door open to a wide variety of other fields, including doing math on the complex plane, polar functions, and many applications in calculus and other fields.

SOH - CAH - TOA

Shhhhh... mathematicians don't like this too much, but for most of you, about 80% of all you'll ever need to know about trig. is in that mnemonic: SOH-CAH-TOA.

Most of the trigonometry you'll ever need to do involves right triangles, though we'll generalize this to any triangle, and do a lot of other interesting things later. Take a look at the right triangles on the right so we can be on the same page with our terms.

Right triangles have a right angle, of course, and a hypotenuse across from that. We'll define an angle, θ (the Greek letter "theta"), and our choice of θ will determine the labels of the other two sides. The opposite side (o) is opposite angle θ, and the side that is a part of theta will be called the adjacent side (a).

Example 1

Calculate the measures of the missing sides and angles of a right triangle.

Solution: The three trig. functions allow us to determine everything there is to know about a right triangle starting with very little initial information. In the triangle below, only one side and one angle (and the right angle) are known.

The angle (θ = 25˚) opens onto the opposite side and the adjacent side is part of that angle. The hypotenuse is the unique side of a right triangle; it is not a part of the right angle.

That's plenty of information, though, to find all of the missing information.

To begin, just write down one of the trigonometric relationships — let's choose sine — and rearrange it to find the missing quantity:

Trig. functions & your calculator

Almost any calculator or calculator application (e.g. on your phone) will be able to calculate sine, cosine and tangent functions, and a few other trig. functions that we'll cover in due time. Calculators make the trig. functions act like "black boxes" – you just key in your angle and you get its sine, cosine or tangent.

You'll need to be careful about the units. Your calculator can use two units of angle measure: Degrees and Radians. Make sure you're working with the right units. Learn more about angle measurements here.

Example 2

Calculate all of the missing angles and sides of this triangle. This time the hypotenuse is one of the missing sides.

Solution: We begin by writing down one of the trigonometric functions. The 18 cm side of this triangle is the side adjacent to the 34˚ angle, so let's use cosine. Here we'll write out the definition of cosine and plug in the side and angle we have, then multiply both sides by h and divide both by cos(34˚).

$$ \begin{align} cos(34˚) = \frac{18}{h} \color{#E90F89}{\longrightarrow} h &= \frac{18}{cos(34˚)} \\[5pt] h &= 21.7 \; cm \end{align}$$

We can find the length of the opposite side, o, using the tangent function in a similar way:

$$ \begin{align} tan(34˚) = \frac{o}{18} \color{#E90F89}{\longrightarrow} o &= 18 \cdot tan(34˚) \\[5pt] o &= 12.1 \; cm \end{align}$$

Practice problems 1

Try to find all angle and side measurements of the three triangles below using SOH-CAH-TOA trig. Roll over the problems to see the answers.

Example 3

In the scenario below, a surveyor can calculate the height of a building by measuring the angle of elevation from her position to the top of the building, as long as the distance to the building (100 ft. in this case) is known. Calculate the height of the building.

Solution: The surveyor's lines-of-sight (dashed red lines) form a right triangle: The straight-line distance to the building, the height of the building (minus the 5 ft. from the ground to the surveyor's instrument) and the hypotenuse. To find the length of the opposite side, which we'll call o, we use the tangent function:

$$tan(19.3˚) = \frac{o}{100}$$

Here I'm calling o the height of the building above our 5 ft. mark. We'll have to add that 5 ft. to the total before we finish. We can rearrange to solve for o, multiplying by 100 ft. on both sides:

$$o = 100 \cdot tan(19.3˚) = 35$$

So the height of the building is that 35 ft. plus the 5 ft. height of the instrument, or 40 ft.

Interesting fact: The tangent function is approximately equal to the line y = x between, say, ±π/6. Here are overlapping graphs to show this:

The unit circle – the fundamental basis for trigonometry

Click/tap on the image to download a copy.

Unit circle learning tool

The special triangles

Two common triangles you should memorize.

These two essential right triangles, their angle measures and the ratios of their side lengths will pop up time and again throughout your studies and work in math and science. Knowing how to jot them down and use them, and from where the measurements came, will be invaluable to you. Study them carefully and memorize their proportions.

The 45-45-90 triangle

The 45-45-90 triangle is just what it sounds like, a right triangle that has two 45˚ angles, and is thus an isosceles triangle. We'll work with just symbols for the lengths of the sides, like this:

Now the Pythagorean theorem has to hold for this right triangle, so

$$^2 + x^2 = r^2$$

Now combine the like terms on the left,

$$2x^2 = r^2$$

and let's solve for x (you'll see why in a minute), by first dividing both sides by 2.

$$x^2 = \frac{r^2}{2}$$

Now to solve for x, take the square root of both sides:

$$x = \sqrt{\frac{r^2}{2}}$$

The 30-60-90 triangle

The 30-60-90 (degrees) right triangle is another frequently-encountered triangle that's worth knowing a lot about.

It isn't quite as easy to find an expression for the side lengths x and y, unless we combine two of these triangles to form an equilateral triangle like this:

Notice here that I've replaced x from the original figure with r/2. We know that the gray vertical line both bisects the top 60˚ angle (by construction) and bisects the base of the triangle. For a proof the the angle bisector of an isosceles triangle also bisects the opposite side, click here.

Now we can set up the Pythagorean theorem for one of the two 30-60-90 triangles shown:

$$r^2 = y^2 + \left( \frac{r}{2} \right)^2$$

Trigonometry – trigonometric functions

Now we want to use the unit circle to create the three major trigonometric functions,

• $f(x) = sin(x)$
• $f(x) = cos(x)$
• $f(x) = tan(x)$

We'll do that by drawing a series of triangles, both 45-45-90 and 30-60-90, inside of the unit circle, and calculating the trig function values.

Let's start at 30˚ = θ/6. Notice that we can use that triangle to calculate all three trig functions. We'll set those aside as we work around the circle with triangles, and look at all of the values together at the end of the trip.

Now we really should work our way around the circle in increments of 30˚, but I'm going to skip around a bit and hope you'll get the idea. Next is a 135˚ angle, or 45˚ past 90˚, working counterclockwise (as usual) around the unit circle.

One quirk of the unit circle is that the length of the radius (the hypotenuses of all of our triangles) is always 1, but the other triangle sides can have negative or positive values. That's how certain sine, cosine or tangent values end up negative.

Trig functions are periodic

Here's a table of trig function values at key angles in one round trip of the unit circle.

The trigonometric functions are repetitive, or cyclic, and we usually refer to them as periodic, meaning that they repeat the same basic pattern predictably.

Graphs of the trig functions

Graphs of the sine (black) and cosine (magenta) functions are drawn below as f(θ) vs. θ, for θ between 0 and 4π - twice around the unit circle. Notice that these are periodic functions. Every 2π radians, they repeat, and that goes on infinitely in both positive and negative directions.

All of the points in the table above are plotted on this graph. Be sure to think about these graphs for a while.

Transformations of the trig functions

In order to model periodic behavior using sines and cosines, we need, as usual, some transformations to translate and scale our function at will. In our typical way (see functions section), the transformations of f(x) = sin(x) are shown on the right.

When we use trig. functions to model real data later, we'll make a slight modification of this prototype equation in order to keep the units straight.

The transformations of the other trig. functions are analogous.

Vertical stretching

Move the slider to see the effect of the vertical-stretching parameter, A, on our sine function.

If this function represented a wave like a sound or light wave, the vertical distance between f(x) = 0 and the maximum (or minimum) value of the sine function would be the amplitude, which would be proportional to volume (loudness) of sound or brightness (intensity) of light.

Notice that when A < 0, the function is reflected across the x-axis, as we would expect for any function.

Vertical and horizontal translation

Vertical translation of a periodic function is performed just like on any other function: Simply add a constant number (k) to the value of the function for each point in the domain. Vertical translation is used in modeling to move the average value of the function up or down. For example, to model a tide that fluctuates by ± 3 feet from an average depth of 10 feet, we'd want a sine or cosine function (turns out it doesn't matter which), that oscillates between 7 and 13 feet. The function would read f(some stuff) + 10.

Reciprocal trig functions: The tangent function

The tangent of an angle (θ) is the ratio of the sine of that angle to its cosine: tan(θ) = sin(θ)/cos(θ). Because cos(θ) can be zero, the graph of the tangent function will have vertical asymptotes, as shown on the right. The tangent of odd multiples of π/2 is infinite because cos(π/2) (and odd multiples of π/2) is zero.

This has an interesting consequence if you are a rock climber or a linesman (person who strings cable, e.g. electric wires, from pole to pole. You'll find an explanation here.

There are three other trigonometric functions left to explore, and like the tangent, all are ratios of other trig. functions, so we can expect asymptotic behavior there, too.

All six trigonometric functions

Finally, let's define the three other trig. functions. The reciprocal of the sine function (not to be confused with the inverse) is called the cosecant (csc) function. The reciprocal of the cosine is (paradoxically, I know) the secant (sec) and the reciprocal of the tangent is the cotangent (cot). csc(θ), sec(θ) and cot(θ) are defined below.

Graphs of the Trigonometric Functions

Several cycles of each of the six trig functions are plotted below. Note that the sin(x) and cos(x) graphs are vertically expanded by a factor of ten compared to the other graphs. This means that the sine graph would fit in between the U-shaped graphs (they are not parabolas!) in the csc(x) graph, same for the cos(x) and sec(x) graphs. Except for sine and cosine,

Practice problems 2

For problems 1-4, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.

The small-angle approximation

Here's an interesting problem and an opportunity to show how approximations can be good and save a lot of time. Consider a boat out on the ocean, a distance d from a lighthouse of known height – in this case 25 m. We'll explore two ways of calculating the distance to the lighthouse. Our measurement from the boat is the angle of elevation between the horizontal (sea surface) and the top of the lighthouse, 2˚.

First, let's do this in the straightforward way using the tangent function. First let's convert the angle to radians, because we'll use it that way later:

$$2˚ \left( \frac{\pi \, \text{rad}}{180˚} \right) = 0.03491 \, \text{rad}$$

The tangent of the angle is

$$tan(0.03491 \, \text{rad}) = \frac{25 \, m}{d},$$

so the distance is

$$d = \frac{25 \, m}{tan(0.03491 \, \text{rad})} = 715.836 \, m$$

Now let's do that in a different way. Notice that the circle of radius d is shown, and that, to a good approximation, the part of that circle that crosses through the lighthouse is almost a line. That is, if we take a small enough part of any curve, it's nearly a line.

Video examples

1. Counting around the unit circle

You should know how to label the most-frequently-used angles around the unit circle (circle of radius r = 1), both in degrees and radians. It's not that difficult if you just think of it as counting around the circle in different increments.

Minutes of your life: 3:25

3. Using basic right-triangle trigonometry

Here are two examples of how we use SOH-CAH-TOA trig. to solve for the missing sides of a right triangle. This is a basic skill in trigonometry that you should master.

Minutes of your life: 4:48 (two examples)