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: 20


$sin~\theta=\frac{opp}{hyp}=\frac{4\sqrt {15}}{17}$ $cos~\theta=\frac{adj}{hyp}=\frac{7}{17}$ $tan~\theta=\frac{opp}{adj}=\frac{4\sqrt {15}}{7}$ $cot~\theta=\frac{adj}{opp}=\frac{7\sqrt {15}}{60}$ $csc~\theta=\frac{hyp}{opp}=\frac{17\sqrt {15}}{60}$

Work Step by Step

$sec~\theta=\frac{hyp}{adj}$ $\frac{17}{7}=\frac{hyp}{adj}$ A right triangle with hypotenuse equal to $17$ and with the adjacent side of $\theta$ equal to $7$ has secant equal to $\frac{17}{7}$ Use the pythagorean theorem to find the opposite side of $\theta$ $17^2=7^2+opp^2$ $opp^2=289-49=240$ $opp=4\sqrt {15}$ $sin~\theta=\frac{opp}{hyp}=\frac{4\sqrt {15}}{17}$ $cos~\theta=\frac{adj}{hyp}=\frac{7}{17}$ $tan~\theta=\frac{opp}{adj}=\frac{4\sqrt {15}}{7}$ $cot~\theta=\frac{adj}{opp}=\frac{7}{4\sqrt {15}}=\frac{7\sqrt {15}}{60}$ $csc~\theta=\frac{hyp}{opp}=\frac{17}{4\sqrt {15}}=\frac{17\sqrt {15}}{60}$
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