Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 13

Answer

b = $\sqrt 13$ $\sin$B = $\frac{\sqrt 13}{7}$ $\cos$B = $\frac{6}{7}$ $\tan$B = $\frac{\sqrt 13}{6}$ $\csc$B = $\frac{7\sqrt 13}{3}$ $\sec$B = $\frac{7}{6}$ $\cot$B = $\frac{6\sqrt 13}{13}$

Work Step by Step

a = 6 c = 7 Pythagorean Theorem: c$^{2}$ = a$^{2}$ + b$^{2}$ Therefore: 7$^{2}$ = 6$^{2}$ + b$^{2}$ 49 = 36 + b b$^{2}$ = 49 - 36 b$^{2}$ = 13 b = $\sqrt 13$ Therefore: $\sin$B = $\frac{opposite}{hypotenuse}$ = $\frac{b}{c}$ = $\frac{\sqrt 13}{7}$ $\cos$B = $\frac{adjacent}{hypotenuse}$ = $\frac{6}{7}$ $\tan$B = $\frac{opposite}{adjacent}$ =$\frac{\sqrt 13}{6}$ $\csc$B = $\frac{hypotenuse}{opposite}$ = $\frac{7}{\sqrt 13}$ = $\frac{7\sqrt 13}{3}$ $\sec$B = $\frac{hypotenuse}{adjacent}$ =$\frac{7}{6}$ $\cot$B = $\frac{adjacent}{opposite}$ = $\frac{6}{\sqrt 13}$ = $\frac{6\sqrt 13}{13}$
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