Algebra and Trigonometry 10th Edition

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

Chapter 6 - Chapter Test - Page 503: 5

Answer

$sin~\theta=\frac{3\sqrt {10}}{10}$ $cos~\theta=-\frac{\sqrt {10}}{10}$ $tan~\theta=-3$ $cot~\theta=-\frac{1}{3}$ $sec~\theta=-\sqrt {10}$ $csc~\theta=\frac{\sqrt {10}}{3}$

Work Step by Step

Distance from the origin to the given point: $d=\sqrt {(-2)^2+6^2}=\sqrt {40}=2\sqrt {10}$ $sin~\theta=\frac{opp}{hyp}=\frac{6}{2\sqrt {10}}=\frac{3\sqrt {10}}{10}$ $cos~\theta=\frac{adj}{hyp}=\frac{-2}{2\sqrt {10}}=-\frac{\sqrt {10}}{10}$ $tan~\theta=\frac{opp}{adj}=\frac{6}{-2}=-3$ $cot~\theta=\frac{adj}{opp}=\frac{-2}{6}=-\frac{1}{3}$ $sec~\theta=\frac{hyp}{adj}=\frac{2\sqrt {10}}{-2}=-\sqrt {10}$ $csc~\theta=\frac{hyp}{opp}=\frac{2\sqrt {10}}{6}=\frac{\sqrt {10}}{3}$
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