Skip to main content
\( \newcommand{\lt}{ < } \newcommand{\gt}{ > } \newcommand{\amp}{ & } \)

SectionB.1Right Triangles and Trigonometry

SubsectionB.1.1Right Triangles

We already reviewed the idea that similar triangles have equal ratios of corresponding sides. This is at the heart of trigonometry using right triangles. An important fact from geometry is that any two triangles that have equal angles are similar. A right triangle, by definition, has one angle that is perpendicular or 90 degrees. Because the sum of angles in a triangle always add to 180 degrees, it really only takes one of the other angles to establish similarity.

We consider an acute angle \(\theta\) (less than 90 degrees). A right triangle with base angle \(\theta\) in standard position (with the base horizontal and the angle on the left) is shown in the diagram below. The legs that join at right angles are identified as being adjacent (adj) to the angle or opposite (opp) to the angle, while the side opposite the right angle is the hypotenuse (hyp).

<<SVG image is unavailable, or your browser cannot render it>>

FigureB.1.1An illustration of similar right triangles with base angle \(\theta\) in standard position.

For any right triangle with the same base angle, the ratios of these three sides are always going to found in the exact proportions. These proportions define the six trigonometric values of the angle: \begin{align*} \sin \theta & = \frac{\mathrm{opp}}{\mathrm{hyp}} & \hbox{(sine)}\\ \cos \theta & = \frac{\mathrm{adj}}{\mathrm{hyp}} & \hbox{(cosine)}\\ \tan \theta & = \frac{\mathrm{opp}}{\mathrm{adj}} & \hbox{(tangent)}\\ \sec \theta & = \frac{\mathrm{hyp}}{\mathrm{adj}} & \hbox{(secant)}\\ \cot \theta & = \frac{\mathrm{adj}}{\mathrm{opp}} & \hbox{(cotangent)}\\ \csc \theta & = \frac{\mathrm{hyp}}{\mathrm{opp}} & \hbox{(cosecant)}. \end{align*} The first three proportions are often introduced using a mnemonic ``SOH-CAH-TOA" where the three letters correspond to the first letter of the proportion name (Sine), the numerator side (Opposite) and denominator side (Hypotenuse).

It is sometimes useful to think of drawing special right triangles where one of the sides has unit length (length=1). When the hypotenuse has unit length, the triangle involves sine (opposite) and cosine (adjacent). When the adjacent leg has unit length, the triangle has sides with lengths given by tangent (opposite) and secant (hypotenuse). When the opposite leg has unit length, the triangle has sides with lengths given by cotangent (adjacent) and cosecant (hypotenuse). When I realized this simple fact, the naming of the proportions made much more sense.

Unit right triangles showing the six fundamental trigonometric values.
FigureB.1.2Illustration of the three similar unit right triangles with angle \(\theta\).
Proof

SubsectionB.1.2Special Right Triangles

There are two right triangles that often appear in problems because they are geometrically simple. One of these is the isosceles right triangle where the base angle is exactly half of a right angle (\(\theta = 45^\circ\)). This triangle is often called a 45–45–90 right triangle. The other simple triangle comes from dividing an equilateral triangle in half, so that the base angle is either \(30^\circ\) or \(60^\circ\), and is called a 30–60–90 right triangle.

<<SVG image is unavailable, or your browser cannot render it>>

FigureB.1.4Illustration of 45–45–90 and 30–60–90 right triangles.

It is handy to be able to reproduce these two triangles as quickly as possible. The key is to remember how the triangles were created. For the 45–45–90 triangle, draw an isosceles right triangle and label the lengths of both legs as 1. To find the length of the hypotenuse $h$, use the Pythagorean theorem: \begin{gather*} 1^2 + 1^2 = h^2 \\ h^2 = 2 \\ h = \sqrt{2} \end{gather*} For the 30–60–90 triangle, recall that this is exactly half of an equilateral triangle. The hypotenuse will have length 1 while the leg opposite the 30 degrees is exactly half that length. Now use the Pythagorean theorem to find the length of the other leg $b$ which bisected the triangle: \begin{gather*} (\frac{1}{2})^2 + b^2 = 1^2 \\ \frac{1}{4} + b^2 = 1 \\ b^2 = \frac{3}{4} \\ b = \frac{\sqrt{3}}{2} \end{gather*}

Once you have the triangles drawn with lengths identified, you can use the triangles to find the proportions that define the trigonometric values. I strongly discourage trying to memorize this table. Learn how the table was created, and that will reinforce the more general principles and ultimately require less mental effort to recall.

\(\theta\) \(30^{\circ}\) \(45^{\circ}\) \(60^{\circ}\)
\(\sin \theta = \frac{\hbox{opp}}{\hbox{hyp}}\) \(\sin 30^{\circ} = \frac{1}{2}\) \(\sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\) \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\)
\(\cos \theta = \frac{\hbox{adj}}{\hbox{hyp}}\) \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\) \(\cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\) \(\cos 60^{\circ} = \frac{1}{2}\)
\(\tan \theta = \frac{\hbox{opp}}{\hbox{adj}}\) \(\tan 30^{\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\) \(\tan 45^{\circ} = 1\) \(\tan 60^{\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\)

SubsectionB.1.3Examples

The first example illustates how you can use two known lengths for a triangle to find the trigonometric values for the angle.

ExampleB.1.5

Suppose a right triangle with a base angle \(\theta\) has a base of length 3 and a height of length 5. What are the trigonometric values associated with \(\theta\)?

Solution

Our second example illustrates how knowing the trigonometric values associated with an angle allow us to determine the lengths of sides. This occurs a lot in physics in the context of decomposing a force or velocity into two perpendicular components.

ExampleB.1.6

A golf ball is launched at a speed of 80 ms and at an angle of \(40^\circ\) from the ground. What are the horizontal and vertical components of the ball's velocity?

Solution