Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set: 4

Answer

sin θ = $\frac{8}{17}$ cos θ = $\frac{15}{17}$ tan θ =$\frac{8}{15}$ csc θ = $\frac{17}{8}$ sec θ = $\frac{17}{15}$ cot θ = $\frac{15}{8}$

Work Step by Step

$a^{2}$ + $b^{y2}$ = $c^{2}$ given b= 15, c= 17 $a^{2}$ + $15^{2}$ = $17^{2}$ $a^{2}$ = 289 - 225 $a^{2}$ = 64 a = $\sqrt {64}$ = 8 sin θ = $\frac{opposite}{hypotenuse}$ = $\frac{8}{17}$ cos θ = $\frac{adjacent}{hypotenuse}$ = $\frac{15}{17}$ tan θ = $\frac{opposite}{adjacent}$ = $\frac{8}{15}$ csc θ = $\frac{1}{sin θ}$ = $\frac{17}{8}$ sec θ = $\frac{1}{cos θ}$ = $\frac{17}{15}$ cot θ = $\frac{1}{tan θ}$ = $\frac{15}{8}$
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.