The word trigonometry comes from the Greek for "triangle measure", and that's a pretty good description, but trig (we call it "trig") goes way beyond that. Learning 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.
Most of trigonometry is based on the geometry of the right triangle and the ratios of the lengths of its sides.
If one angle (labeled with the Greek letter θ here) is our "angle of interest,", then the side adjacent to it is labeled $a$ for adjacent,
the side opposite is labeled $o$, and the hypotenuse, opposite from the right angle, is always a unique side in a right triangle. Remember that all right triangles are similar triangles.
If you're thinking about more advanced courses in math and science, in high school or in college, learn the basics of right-triangle trigonometry well. You won't regret it.
Did you know you couldn't get an MRI image of your injured knee if it weren't for trigonometry? Trig. is everywere.
Shhhhh the ... 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).
In trigonometry — and many other subjects in math and science — we often use Greek letters because we'd run out of letters from the English alphabet. We use Greek letters to stand for angles in trigonometry. The most common are:
The sine of an angle of a right triangle is the ratio of the opposite side to the hypotenuse
$$sin(\theta) = \frac{o}{h}$$
The cosine of an angle of a right triangle is the ratio of the adjacent side to the hypotenuse
$$cos(\theta) = \frac{a}{h}$$
The tangent of an angle of a right triangle is the ratio of the opposite side to the adjacent side.
$$tan(\theta) = \frac{o}{a}$$
The trigonometric functions can't take on just any values — they have limits. Look at this right triangle, for example:
In this triangle, the angle is very small, making the opposite side small. That means that the sine of $\theta$ will be small and the cosine of $\theta$ will be close to 1.
$$ \begin{align} sin(\theta) &= \frac{o}{h} \; \rightarrow \frac{0}{h} \; \rightarrow 0 \\[5pt] cos(\theta) &= \frac{a}{h} \; \rightarrow \frac{a}{h} \; \rightarrow 1 \\[5pt] tan(\theta) &= \frac{o}{a} \; \rightarrow \frac{0}{a\approx h} \; \rightarrow 0 \end{align}$$
Notice that as the angle $\theta$ gets close to zero, the sine ratio gets closer to zero, and is exactly zero when $\theta = 0$. Likewise, as $\theta$ gets close to zero, the lengths of the adjacent side and the hypotenuse approach being the same length, so the cosine ratio approaches 1. Finally, as $\theta$ approaches zero, the tangent ratio, opposite divided by adjacent, also approaches zero. These are hard limits on our three trig functions as $\theta \; \rightarrow 0$.
Now if $\theta$ gets close to 90˚, we have a triangle that looks like this one. Now we see that as $\theta \; \rightarrow \; 90^{\circ}$, the adjacent side gets smaller and the opposite side gets closer to the lenth of the hypotenuse. So we have these trends:
$$ \begin{align} sin(\theta) &= \frac{o}{h} \; \rightarrow \frac{o \approx h}{h} \; \rightarrow 1 \\[5pt] cos(\theta) &= \frac{a}{h} \; \rightarrow \frac{0}{h} \; \rightarrow 0 \\[5pt] tan(\theta) &= \frac{o}{a} \; \rightarrow \frac{o \approx h}{0} \; \rightarrow \infty \end{align}$$
As angle $\theta$ gets close to 90˚, the sine function gets closer to 1, and is one when $\theta = 90^{\circ}$. Likewise, the cosine function gets closer to zero because the adjacent side gets smaller and smaller. Finally, the tangent becomes infinite because its denominator, the adjacent side, approaches zero.
$$ \begin{align} sin(\theta): \; D: (-\infty, \infty) &\phantom{000} R: (-1, 1) \\[5pt] cos(\theta): \; D: (-\infty, \infty) &\phantom{000} R: (-1, 1) \\[5pt] tan(\theta): \; D: \left( \dots -\frac{\pi}{2}, \frac{\pi}{2}, \dots \right) &\phantom{000} R: (-\infty, \infty) \tag{*} \end{align}$$
* The tangent function has a more complicated domain than we can explain in this section. Don't worry about it right now; we'll get there.
Calculate the measures of the missing sides and angles of a right triangle.
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:
$$sin(\theta) = \frac{o}{h} \color{#E90F89}{\longrightarrow} o = h \cdot sin(\theta)$$
So the side opposite the 25˚ angle has length
$$o = 10 \cdot sin(25˚) = 4.2 \; cm$$
There is now enough information to find the third (adjacent) side using the Pythagorean theorem, but let's use the cosine, just for practice:
$$cos(\theta) = \frac{a}{h} \color{#E90F89}{\longrightarrow} a = h \cdot cos(\theta)$$
And the length of the adjacent side is
$$a = 10 \cdot cos(25˚) = 9.1 \; cm$$
You can confirm this for yourself using the Pythagorean theorem.
Lastly, we can easily find the degree-measure of the missing angle,
$$\phi = 180˚ - 25˚ - 90˚ = 55˚$$
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.
Calculate all of the missing angles and sides of this triangle. This time the hypotenuse is one of the missing sides.
$$ \begin{align} cos(34˚) = \frac{18}{h} \color{#E90F89}{\longrightarrow} h &= \frac{18}{cos(34˚)} \\[5pt] h &= 21.7 \; cm \end{align}$$
We can find the length of the opposite side, o, using the tangent function in a similar way:
$$ \begin{align} tan(34˚) = \frac{o}{18} \color{#E90F89}{\longrightarrow} o &= 18 \cdot tan(34˚) \\[5pt] o &= 12.1 \; cm \end{align}$$
So we have sides of 18 cm, 12.1 cm and 21.7 cm. We can easily check whether these are correct using the Pythagorean theorem:
$$ \begin{align} 21.7^2 &\stackrel{?}{=} 18^2 + 12.1^2 \\[5pt] 470.89 &= 324 + 146.41 \\[5pt] &= 470.41 \end{align}$$
The small amount of error, 0.48 cm, is just due to rounding off the earlier answers. Finally, the missing angle is just the complement of the 34˚ angle (the right angle accounts for the other 90˚ of a right triangle):
$$90˚ - 34˚ = 56˚$$
So we know everything there is to know about that triangle, even though we originally only knew the measures of two angles and one side.
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.
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.
$$tan(19.3˚) = \frac{o}{100}$$
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:
This means that tan(π/6) is approximately π/6. Now π/6 is about 15˚, so for angles of elevation between about ±15˚, we can make a pretty good estimate of the height of objects without even calculating the tangent ... But ... this only works if we use radians instead of degrees. In this problem, the approximation would have yielded a building height of just under 39 ft. Not too bad.
For problems 1-4, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.
The missing angle is $x = 180 - 90 - 39 = 51^{\circ}$.
The length of the hypotenuse (h) can be obtained using
$$ \begin{align} sin(39^{\circ}) &= \frac{2}{h} \\[5pt] h &= \frac{2}{sin(39^{\circ})} = 3.2 \, \text{cm} \end{align}$$
The adjacent side is
$$ \begin{align} tan(39^{\circ}) &= \frac{2}{a} \\[5pt] a &= \frac{2}{tan(39^{\circ})} = 2.5 \, \text{cm} \end{align}$$
The missing angle is $x = 180 - 90 - 15 = 75^{\circ}$.
The length of the opposite side (o) can be obtained using
$$ \begin{align} sin(15^{\circ}) &= \frac{o}{12} \\[5pt] o &= 12 \cdot sin(15^{\circ}) = 3.1 \, \text{m} \end{align}$$
We can get the adjacent side like this:
$$ \begin{align} cos(15^{\circ}) &= \frac{a}{12} \\[5pt] a &= 12 \cdot cos(15^{\circ}) = 11.6 \, \text{m} \end{align}$$
The missing angle is $x = 180 - 90 - 63 = 27^{\circ}$.
The length of the hypotenuse (h) can be obtained using
$$ \begin{align} sin(63^{\circ}) &= \frac{1.04}{h} \\[5pt] h &= \frac{1.04}{sin(63^{\circ})} = 1.17 \, \text{m} \end{align}$$
The adjacent side is
$$ \begin{align} tan(63^{\circ}) &= \frac{1.04}{a} \\[5pt] a &= \frac{1.04}{tan(63^{\circ})} = 0.53 \, \text{m} \end{align}$$
The missing angle is $x = 180 - 90 - 41 = 49^{\circ}$.
The length of the opposite side (o) can be obtained using
$$ \begin{align} sin(41^{\circ}) &= \frac{o}{18.5} \\[5pt] o &= 18.5 \cdot sin(41^{\circ}) = 12.1 \, \text{in.} \end{align}$$
We can get the adjacent side like this:
$$ \begin{align} cos(41^{\circ}) &= \frac{a}{18.5} \\[5pt] a &= 18.5 \cdot cos(41^{\circ}) = 13.9 \, \text{in.} \end{align}$$
A surveyor knows the depth of the canyon and can measure the angle shown. Calculate the distance from one side of the canyon to the other.
Here's the basic geometry of the problem:
Our task is to find the length of the adjacent side of that triangle, which suggests the tangent:
$$ \begin{align} tan(14.3^{\circ}) &= \frac{250}{a} \\[5pt] a &= \frac{250}{tan(14.3^{\circ})} = 981 \, \text{m} \end{align}$$
The canyon is about 981 m across.
Roof trusses are what hold up the roof and keep the walls parallel in some houses. They distribute any forces (mostly the downward forces of gravity, but also wind forces) evenly among all of the wooden members ("chords") that make them up. Using the angles and measurements given, determine the lengths A and B shown in the figure.
Here's the basic geometry of the problem:
To find angle $\theta$ we subtract: $180 - 55 = 2\theta = 125^{\circ}$, so $\theta = 62.5^{\circ}$
We can use the sine function to find the $B$ length:
$$ \begin{align} sin(62.5^{\circ}) &= \frac{4}{B} \\[5pt] B &= \frac{4}{sin(62.5^{\circ})} = 4.51 \, \text{ft.} \end{align}$$
To find $A/2$ we can either use the Pythagorean theorem or the tangent function. Given that we're here to learn trig, let's use the tangent.
$$ \begin{align} tan(62.5^{\circ}) &= \frac{4}{A/2} \\[5pt] \frac{A}{2} &= \frac{4}{tan(62.5^{\circ})} \\[5pt] A &= 2 \cdot \frac{4}{tan(62.5^{\circ})} \\[5pt] &= 4.16 \, \text{ft.} \end{align}$$
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.
If we approximate that the length of the arc is about the length of the lighthouse, and recall that arc length (s) when using radians is
$$s = r \theta,$$
Then we can solve for the radius of the circle (which is the distance, d, here) like this:
$$r = \frac{s}{\theta} = \frac{25 \, m}{0.03491 \, \text{rad}} = 716.127 \, m$$
That's a difference of only about 0.04%, not to bad. When the angles are small, the small-angle approximation is a good one. This may not seem like such a big deal, but it turns out that when we calculate trig. function values on a computer, it takes a lot of time, relatively-speaking. The number of multiplications (or divisions) is the key. Our approximation just takes one division while calculating a tangent value can take several. In a computer program that might execute millions of these calculations, that can be a significant savings – if you can afford the small loss of precision.
You'll make your journey through trigonometry much easier if you memorize everything needed to derive the trig functions of 30˚, 45˚ and 60˚ angles. Here's how to do it.
Minutes of your life: 3:42
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)
A mnemonic (nee·mon'·ick) is a word or phrase designed to help a person remember something. An example would be the pseudo-word "ROYGBIV" or the phrase "Rogers of York Gave Battle in Vain." Both are designed to help us remember the colors of the visible spectrum: red, orange, yellow, green, blue, indigo & violet.
alpha | Α | α |
beta | Β | β |
gamma | Γ | γ |
delta | Δ | δ |
epsilon | Ε | ε |
zeta | Ζ | ζ |
eta | Η | η |
theta | Θ | θ |
iota | Ι | ι |
kappa | Κ | κ |
lambda | Λ | λ |
mu | Μ | μ |
nu | Ν | ν |
xi | Ξ | ξ |
omicron | Ο | ο |
pi | Π | π |
rho | Ρ | ρ |
sigma | Σ | σ |
tau | Τ | τ |
upsilon | Υ | υ |
phi | Φ | φ |
chi | Χ | χ |
psi | Ψ | ψ |
omega | Ω | ω |
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