Trigonometry (10th Edition)

Published by Pearson

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises - Page 51: 17

Answer

b = $\sqrt 3$ $\sin$$B = \frac{\sqrt 3}{2} \cos$$B$ = $\frac{1}{2}$ $\tan$$B = \frac{\sqrt 3}{1} \csc$$B$ = $\frac{2\sqrt 3}{3}$ $\sec$$B = 2 \cot$$B$ =$\frac{1}{\sqrt 3}$ = $\frac{\sqrt 3}{3}$

Work Step by Step

a = 1 c = 2 Pythagorean Theorem: c$^{2}$ = a$^{2}$ + b$^{2}$ 2$^{2}$ = 1$^{2}$ + b$^{2}$ 4 = 1 + b$^{2}$ b$^{2}$ = 3 b = $\sqrt 3$ $\sin$$B = \frac{opposite}{hypotenuse} = \frac{b}{c} = \frac{\sqrt 3}{2} \cos$$B$ = $\frac{adjacent}{hypotenuse}$ = $\frac{a}{c}$ = $\frac{1}{2}$ $\tan$$B = \frac{opposite}{adjacent} = \frac{b}{a} = \frac{\sqrt 3}{1} \csc$$B$ = $\frac{hypotenuse}{opposite}$ = $\frac{c}{b}$ = $\frac{2}{\sqrt 3}$ = $\frac{2\sqrt 3}{3}$ $\sec$$B = \frac{hypotenuse}{adjacent} = \frac{c}{a} = \frac{2}{1} = 2 \cot$$B$ = $\frac{adjacent}{opposite}$ = $\frac{a}{b}$ = $\frac{1}{\sqrt 3}$ = $\frac{\sqrt 3}{3}$

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