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: 15


$sin~\theta=\frac{opp}{hyp}=\frac{8}{17}$ $tan~\theta=\frac{opp}{adj}=\frac{8}{15}$ $cot~\theta=\frac{adj}{opp}=\frac{15}{8}$ $sec~\theta=\frac{hyp}{adj}=\frac{17}{15}$ $csc~\theta=\frac{hyp}{opp}=\frac{17}{8}$

Work Step by Step

$cos~\theta=\frac{adj}{hyp}$ $\frac{15}{17}=\frac{adj}{hyp}$ A right triangle with a hypotenuse equal to $17$ and with the adjacent side of $\theta$ equals $15$ has a cosine equal to $\frac{15}{17}$. Use the pythagorean theorem to find the opposite side of $\theta$. $17^2=15^2+opp^2$ $opp^2=289-225=64$ $opp=8$ $sin~\theta=\frac{opp}{hyp}=\frac{8}{17}$ $tan~\theta=\frac{opp}{adj}=\frac{8}{15}$ $cot~\theta=\frac{adj}{opp}=\frac{15}{8}$ $sec~\theta=\frac{hyp}{adj}=\frac{17}{15}$ $csc~\theta=\frac{hyp}{opp}=\frac{17}{8}$
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