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: 14


$\sin\theta=\dfrac{3}{4}$ $\cos\theta=\dfrac{\sqrt 7}{4}$ $\tan\theta=\dfrac{3\sqrt 7}{7}$ $\cot\theta=\dfrac{\sqrt 7}{3}$ $\sec\theta=\dfrac{4\sqrt 7}{7}$ $\csc\theta=\dfrac{4}{3}$

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: $h=4$ $o=3$ Determine the adjacent side, using the Pythagorean Theorem: $h^2=o^2+a^2$ $4^2=3^2+a^2$ $16=9+a^2\\ a^2=7$ $a=\pm\sqrt{7}$ Since $a$ is never negative. then $a=\sqrt7$. Determine the $6$ trigonometric functions of angle $\theta$: $\sin\theta=\dfrac{o}{h}=\dfrac{3}{4}$ $\cos\theta=\dfrac{a}{h}=\dfrac{\sqrt 7}{4}$ $\tan\theta=\dfrac{o}{a}=\dfrac{3}{\sqrt 7}=\dfrac{3\sqrt 7}{7}$ $\cot\theta=\dfrac{a}{o}=\dfrac{\sqrt 7}{3}$ $\sec\theta=\dfrac{h}{a}=\dfrac{4}{\sqrt 7}=\dfrac{4\sqrt 7}{7}$ $\csc\theta=\dfrac{h}{o}=\dfrac{4}{3}$
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