Algebra and Trigonometry 10th Edition

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

Chapter 6 - Review Exercises - Page 500: 32

Answer

$cos~\theta=\frac{1}{2}$ $sin~\theta=\frac{\sqrt 3}{2}$ $tan~\theta=\sqrt 3$ $cot~\theta=\frac{\sqrt 3}{3}$ $sec~\theta=2$ $csc~\theta=\frac{2\sqrt 3}{3}$

Work Step by Step

$hyp=8$ $adj=4$ Use the pythagorean theorem to find the opposite side of $\theta$ $8^2=4^2+opp^2$ $opp^2=64-16=48$ $opp=4\sqrt 3$ $cos~\theta=\frac{adj}{hyp}=\frac{4}{8}=\frac{1}{2}$ $sin~\theta=\frac{opp}{hyp}=\frac{4\sqrt 3}{8}=\frac{\sqrt 3}{2}$ $tan~\theta=\frac{opp}{adj}=\frac{4\sqrt 3}{4}=\sqrt 3$ $cot~\theta=\frac{adj}{opp}=\frac{4}{4\sqrt 3}=\frac{\sqrt 3}{3}$ $sec~\theta=\frac{hyp}{adj}=\frac{8}{4}=2$ $csc~\theta=\frac{hyp}{opp}=\frac{8}{4\sqrt 3}=\frac{2\sqrt 3}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.