## Trigonometry (10th Edition)

Published by Pearson

# Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 18

#### Answer

b = $\sqrt 2$ $\sin$$B = \frac{\sqrt 2}{2} \cos$$B$ = $\frac{\sqrt 2}{2}$ $\tan$$B = 1 \csc$$B$ = $\frac{2\sqrt 2}{2}$ $\sec$$B = \frac{2\sqrt 2}{2} \cot$$B$ = 1

#### Work Step by Step

a = $\sqrt2$ c = 2 Pythagorean Theorem: c$^{2}$ = a$^{2}$ + b$^{2}$ 2$^{2}$ = ($\sqrt2$)$^{2}$ + b$^{2}$ 4 = 2 + b$^{2}$ b$^{2}$ = 2 b = $\sqrt 2$ $\sin$$B = \frac{opposite}{hypotenuse} = \frac{b}{c} = \frac{\sqrt 2}{2} \cos$$B$ = $\frac{adjacent}{hypotenuse}$ = $\frac{a}{c}$ = $\frac{\sqrt 2}{2}$ $\tan$$B = \frac{opposite}{adjacent} = \frac{b}{a} = \frac{\sqrt 2}{\sqrt 2} = 1 \csc$$B$ = $\frac{hypotenuse}{opposite}$ = $\frac{c}{b}$ = $\frac{2}{\sqrt 2}$ = $\frac{2\sqrt 2}{2}$ $\sec$$B = \frac{hypotenuse}{adjacent} = \frac{c}{a} = \frac{2}{\sqrt 2} = \frac{2\sqrt 2}{2} \cot$$B$ = $\frac{adjacent}{opposite}$ = $\frac{a}{b}$ = $\frac{\sqrt 2}{\sqrt 2}$ = 1

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