Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.2 - Right Triangle Trigonometry - 6.2 Exercises - Page 441: 22

Answer

$cos~\theta=\frac{adj}{hyp}=\frac{4\sqrt 5}{9}$ $sin~\theta=\frac{opp}{hyp}=\frac{1}{9}$ $tan~\theta=\frac{opp}{adj}=\frac{1}{4\sqrt 5}=\frac{\sqrt 5}{20}$ $cot~\theta=\frac{adj}{opp}=\frac{4\sqrt 5}{1}=4\sqrt 5$ $sec~\theta=\frac{hyp}{adj}=\frac{9}{4\sqrt 5}=\frac{9\sqrt 5}{20}$

Work Step by Step

$csc~\theta=\frac{hyp}{opp}$ $9=\frac{hyp}{opp}$ Use the pythagorean theorem to find the adjacent side of $\theta$ $9^2=1^2+adj^2$ $adj^2=81-1=80$ $adj=4\sqrt 5$ $cos~\theta=\frac{adj}{hyp}=\frac{4\sqrt 5}{9}$ $sin~\theta=\frac{opp}{hyp}=\frac{1}{9}$ $tan~\theta=\frac{opp}{adj}=\frac{1}{4\sqrt 5}=\frac{\sqrt 5}{20}$ $cot~\theta=\frac{adj}{opp}=\frac{4\sqrt 5}{1}=4\sqrt 5$ $sec~\theta=\frac{hyp}{adj}=\frac{9}{4\sqrt 5}=\frac{9\sqrt 5}{20}$
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