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# Chapter 7: Trigonometry

Difficulty Level: At Grade Created by: CK-12

Here you will learn about trigonometry, the study of triangles. You will use similar triangles to connect the sides and angles of right triangles. You will learn how to use the three trigonometric ratios of tangent, sine, and cosine to find missing sides and angles within right triangles. Then, you will extend your knowledge of the trigonometric ratios. You will see how you can use the sine ratio to find the area of a triangle. You will learn how sine and cosine can be applied to non-right triangles. Finally, you will explore many different applications of trigonometry.

## Chapter Outline

### Chapter Summary

You learned that right triangles with one pair of acute angles congruent are similar, so the ratios between their sides are constant. This led to the development of the tangent ratio, sine ratio, and cosine ratio. Due to the nature of these ratios, the sine and cosine of complementary angles are equal. The tangent of complementary angles are reciprocals.

Inverse trigonometric ratios work backwards, allowing you to find missing angles when you know the ratio between two particular sides of right triangles. The Law of Sines and the Law of Cosines extends the trigonometric ratios to non-right triangles.

The following summarizes the key formulas that were used in this chapter:

• SOH CAH TOA is a mnemonic device to help you remember the three trigonometric ratios:

sinθ=opposite leghypotenusecosθ=adjacent leghypotenusetanθ=opposite legadjacent leg\begin{align*}\sin \theta = \frac{\text{opposite leg}}{\text{hypotenuse}} \quad \cos \theta = \frac{\text{adjacent leg}}{\text{hypotenuse}} \quad \tan \theta = \frac{\text{opposite leg}}{\text{adjacent leg}}\end{align*}

• The Law of Sines: sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\end{align*} (watch out for the SSA case)
• The Law of Cosines: c2=a2+b22abcosC\begin{align*}c^2 = a^2 + b^2 - 2ab \cos C\end{align*}
• The area of a triangle is 12bh\begin{align*}\frac{1}{2}bh\end{align*} or 12absinC\begin{align*}\frac{1}{2}ab \sin C\end{align*}

Aug 27, 2013