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: 16


$\sin\theta=\dfrac{2\sqrt{7}}{7}$ $\cos\theta=\dfrac{\sqrt {21}}{7}$ $\tan\theta=\dfrac{2\sqrt 3}{3}$ $\cot\theta=\dfrac{\sqrt 3}{2}$ $\sec\theta=\dfrac{\sqrt{21}}{3}$ $\csc\theta=\dfrac{\sqrt 7}{2}$

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: $o=2$ $a=\sqrt 3$ Determine the hypotenuse, using the Pythagorean Theorem: $h^2=o^2+a^2$ $h^2=2^2+(\sqrt 3)^2$ $h^2=7$ $h=\pm\sqrt7$ Since $h$ us never negative, then $h=\sqrt{7}$. Determine the $6$ trigonometric functions of angle $\theta$: $\sin\theta=\dfrac{o}{h}=\dfrac{2}{\sqrt 7}=\dfrac{2\sqrt{7}}{7}$ $\cos\theta=\dfrac{a}{h}=\dfrac{\sqrt 3}{\sqrt 7}=\dfrac{\sqrt {21}}{7}$ $\tan\theta=\dfrac{o}{a}=\dfrac{2}{\sqrt 3}=\dfrac{2\sqrt 3}{3}$ $\cot\theta=\dfrac{a}{o}=\dfrac{\sqrt 3}{2}$ $\sec\theta=\dfrac{h}{a}=\dfrac{\sqrt 7}{\sqrt 3}=\dfrac{\sqrt{21}}{3}$ $\csc\theta=\dfrac{h}{o}=\dfrac{\sqrt 7}{2}$
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