Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.1 Right Triangle Trigonometry; Applications - 8.1 Assess Your Understanding - Page 519: 18


$\sin\theta=\dfrac{\sqrt{5}}{5}$ $\cos\theta=\dfrac{2\sqrt 5}{5}$ $\tan\theta=\dfrac{1}{2}$ $\cot\theta=2$ $\sec\theta=\dfrac{\sqrt 5}{2}$ $\csc\theta=\sqrt 5$

Work Step by Step

Let's note: $h$=the hypotenuse $o$=the opposite side of angle $\theta$ $a$=the adjacent side of angle $\theta$ We are given: $a=2$ $h=\sqrt 5$ Determine the opposite side, using the Pythagorean Theorem: $h^2=o^2+a^2$ $(\sqrt 5)^2=o^2+2^2$ $o^2=1$ $o=\pm1$ Since $o$ is never negative, then $o=\sqrt5$. Determine the $6$ trigonometric functions of angle $\theta$: $\sin\theta=\dfrac{o}{h}=\dfrac{1}{\sqrt 5}=\dfrac{\sqrt{5}}{5}$ $\cos\theta=\dfrac{a}{h}=\dfrac{2}{\sqrt 5}=\dfrac{2\sqrt 5}{5}$ $\tan\theta=\dfrac{o}{a}=\dfrac{1}{2}$ $\cot\theta=\dfrac{a}{o}=\dfrac{2}{1}=2$ $\sec\theta=\dfrac{h}{a}=\dfrac{\sqrt 5}{2}$ $\csc\theta=\dfrac{h}{o}=\dfrac{\sqrt 5}{1}=\sqrt 5$
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.