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: The Trigonometric Ratios

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

James Sousa: Introduction to Trigonometric Functions Using Triangles

### Guidance

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

**depending on which**

*opposite***angle is being used.**

*acute*

\begin{align*}& a \ \text{is} \ adjacent \ \text{to} \ \angle B \qquad \ a \ \text{is} \ opposite \ \angle A\\ & b \ \text{is} \ adjacent \ \text{to} \ \angle A \qquad \ b \ \text{is} \ opposite \ \angle B\\ & c \ \text{is the} \ hypotenuse\end{align*}

The three basic trigonometric ratios are called sine, cosine and tangent. For right triangle \begin{align*}\triangle ABC\end{align*}, we have:

**Sine Ratio:** \begin{align*}\frac{opposite \ leg }{hypotenuse} \ \sin A = \frac{a}{c}\end{align*} or \begin{align*}\sin B = \frac{b}{c}\end{align*}

**Cosine Ratio:** \begin{align*}\frac{adjacent \ leg}{hypotenuse} \ \cos A = \frac{b}{c}\end{align*} or \begin{align*}\cos B = \frac{a}{c}\end{align*}

**Tangent Ratio:** \begin{align*}\frac{opposite \ leg}{adjacent \ leg} \ \tan A = \frac{a}{b}\end{align*} or \begin{align*}\tan B = \frac{b}{a}\end{align*}

An easy way to remember ratios is to use SOH-CAH-TOA.

**A few important points:**

- Always
**reduce ratios**(fractions) when you can. - Use the Pythagorean Theorem to find the missing side (if there is one).
- If there is a radical in the denominator,
**rationalize the denominator.**

#### 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 &= c^2\\ 13 &= c\\ \sin A &= \frac{leg \ opposite \ \angle A}{hypotenuse} = \frac{12}{13} && \cos A = \frac{leg \ adjacent \ to \ \angle A}{hypotenuse}= \frac{5}{13},\\ \tan A &= \frac{leg \ opposite \ \angle A}{leg \ adjacent \ to \ \angle 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}\\ \sin B &= \frac{10 \sqrt{2}}{15} = \frac{2 \sqrt{2}}{3} && \cos B = \frac{5}{15}=\frac{1}{3} && \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 30-60-90 triangle. The short leg is 6, \begin{align*}y = 6 \sqrt{3}\end{align*} and \begin{align*}x=12\end{align*}.

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

CK-12 Foundation: The Trigonometric Ratios

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### 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*}

### Explore More

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.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 8.7.