What if you were given a right triangle and told that its sides measure 3, 4, and 5 inches? How could you find the sine, cosine, and tangent of one of the triangle's non-right angles? After completing this Concept, you'll be able to solve for these trigonometric ratios.

### Watch This

CK-12 Foundation: Chapter8SoneCosineTangentA

Watch the parts of the video dealing with the sine, cosine, and tangent.

James Sousa: Introduction to Trigonometric Functions Using Triangles

### Guidance

The word trigonometry comes from two words meaning *triangle* and *measure*. In this lesson we will define three trigonometric (or trig) functions.

**Trigonometry:** The study of the relationships between the sides and angles of right triangles.

In trigonometry, sides are named in reference to a particular angle. The hypotenuse of a triangle is always the same, but the terms **adjacent** and **opposite** depend on which angle you are referencing. A side adjacent to an angle is the leg of the triangle that helps form the angle. A side opposite to an angle is the leg of the triangle that does not help form the angle. We never reference the right angle when referring to trig ratios.

The three basic trig ratios are called, sine, cosine and tangent. At this point, we will only take the sine, cosine and tangent of acute angles. However, you will learn that you can use these ratios with obtuse angles as well.

**Sine Ratio:** For an acute angle \begin{align*}x\end{align*} in a right triangle, the \begin{align*}\sin x\end{align*} is equal to the ratio of the side opposite the angle over the hypotenuse of the triangle. Using the triangle above, \begin{align*}\sin A = \frac{a}{c}\end{align*} and \begin{align*}\sin B = \frac{b}{c}\end{align*}.

**Cosine Ratio:** For an acute angle \begin{align*}x\end{align*} in a right triangle, the \begin{align*}\cos x\end{align*} is equal to the ratio of the side adjacent to the angle over the hypotenuse of the triangle.Using the triangle above, \begin{align*}\cos A = \frac{b}{c}\end{align*} and \begin{align*}\cos B = \frac{a}{c}\end{align*}.

**Tangent Ratio:** For an acute angle \begin{align*}x\end{align*}, in a right triangle, the \begin{align*}\tan x\end{align*} is equal to the ratio of the side opposite to the angle over the side adjacent to \begin{align*}x\end{align*}. Using the triangle above, \begin{align*}\tan A = \frac{a}{b}\end{align*} and \begin{align*}\tan B = \frac{b}{a}\end{align*}.

There are a few important things to note about the way we write these ratios. First, keep in mind that the abbreviations \begin{align*}\sin x, \cos x\end{align*}, and \begin{align*}\tan x\end{align*} are all functions. Second, be careful when using the abbreviations that you still pronounce the full name of each function. When we write \begin{align*}\sin x\end{align*} it is still pronounced *sine,* with a long “\begin{align*}i\end{align*}”. When we write \begin{align*}\cos x\end{align*}, we still say co-sine. And when we write \begin{align*}\tan x\end{align*}, we still say tangent. An easy way to remember ratios is to use the pneumonic SOH-CAH-TOA.

A few important points:

- Always reduce ratios when you can.
- Use the Pythagorean Theorem to find the missing side (if there is one).
- The tangent ratio can be bigger than 1 (the other two cannot).
- If two right triangles are similar, then their sine, cosine, and tangent ratios will be the same (because they will reduce to the same ratio).
- If there is a radical in the denominator, rationalize the denominator.
- The sine, cosine and tangent for an angle are fixed.

#### Example A

Find the sine, cosine and tangent ratios of \begin{align*}\angle A\end{align*}.

First, we need to use the Pythagorean Theorem to find the length of the hypotenuse.

\begin{align*}5^2 + 12^2 & = h^2\\ 13 & = h\end{align*}

So, \begin{align*}\sin A = \frac{12}{13}, \cos A = \frac{5}{13}\end{align*}, and \begin{align*}\tan A = \frac{12}{5}\end{align*}.

#### Example B

Find the sine, cosine, and tangent of \begin{align*}\angle B\end{align*}.

Find the length of the missing side.

\begin{align*}AC^2 + 5^2 & = 15^2\\ AC^2 & = 200\\ AC & = 10 \sqrt{2}\end{align*}

Therefore, \begin{align*}\sin B = \frac{10 \sqrt{2}}{15} = \frac{2 \sqrt{2}}{3}, \cos B = \frac{5}{15} = \frac{1}{3}\end{align*}, and \begin{align*}\tan B = \frac{10 \sqrt{2}}{5} = 2 \sqrt{2}\end{align*}.

#### Example C

Find the sine, cosine and tangent of \begin{align*}30^\circ\end{align*}.

This is a special right triangle, a 30-60-90 triangle. So, if the short leg is 6, then the long leg is \begin{align*}6 \sqrt{3}\end{align*} and the hypotenuse is 12.

\begin{align*}\sin 30^\circ = \frac{6}{12} = \frac{1}{2}, \cos 30^\circ = \frac{6 \sqrt{3}}{12} = \frac{\sqrt{3}}{2}\end{align*}, and \begin{align*}\tan 30^\circ = \frac{6}{6 \sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter8SineCosineTangentB

#### Concept Problem Revisited

The trigonometric ratios for the non-right angles in the triangle above are:

\begin{align*} \sin A = \frac{4}{5}, \cos A = \frac{3}{5}, \tan A = \frac{4}{3}, \sin B = \frac{3}{5}, \cos B = \frac{4}{5},\end{align*} and \begin{align*} \tan B = \frac{3}{4}\end{align*}.

### Vocabulary

** Trigonometry** is the study of the relationships between the sides and angles of right triangles. The legs are called

**or**

*adjacent***depending on which**

*opposite***angle is being used. The three trigonometric (or trig) ratios are**

*acute***,**

*sine***, and**

*cosine***.**

*tangent*### Guided Practice

Answer the questions about the following image. Reduce all fractions.

1. What is \begin{align*}\sin A\end{align*}?

2. What is \begin{align*}\cos A\end{align*}?

3. What is \begin{align*}\tan A\end{align*}?

**Answers:**

1. \begin{align*}\sin A=\frac{16}{20}=\frac{4}{5}\end{align*}

2. \begin{align*} \cos A=\frac{12}{20}=\frac{3}{5}\end{align*}

3. \begin{align*} \tan A=\frac{16}{12}=\frac{4}{3}\end{align*}

### Practice

Use the diagram to fill in the blanks below.

- \begin{align*}\tan D = \frac{?}{?}\end{align*}
- \begin{align*}\sin F = \frac{?}{?}\end{align*}
- \begin{align*}\tan F = \frac{?}{?}\end{align*}
- \begin{align*}\cos F = \frac{?}{?}\end{align*}
- \begin{align*}\sin D = \frac{?}{?}\end{align*}
- \begin{align*}\cos D = \frac{?}{?}\end{align*}

From questions 1-6, we can conclude the following. Fill in the blanks.

- \begin{align*}\cos \underline{\;\;\;\;\;\;\;} = \sin F\end{align*} and \begin{align*}\sin \underline{\;\;\;\;\;\;\;} = \cos F\end{align*}.
- \begin{align*}\tan D\end{align*} and \begin{align*}\tan F\end{align*} are _________ of each other.

Find the sine, cosine and tangent of \begin{align*}\angle A\end{align*}. Reduce all fractions and radicals.

- Explain why the sine of an angle will never be greater than 1.
- Explain why the tangent of a \begin{align*}45^\circ\end{align*} angle will always be 1.
- As the degree of an angle increases, will the tangent of the angle increase or decrease? Explain.