Algebra and Trigonometry 10th Edition

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

Chapter 7 - Review Exercises - Page 552: 17


$\sqrt {25-x^2}=5~cos~θ$

Work Step by Step

$sin^2θ+cos^2θ=1$ $cos^2θ=1-sin^2θ$ $\sqrt {25-x^2}=\sqrt {25-(5~sin^2θ)^2}=\sqrt {25-25~sin^2θ}=\sqrt {25(1-sin^2θ)}=5\sqrt {cos^2θ}=±5~cos~θ$ But, since $0\lt θ\lt\frac{\pi}{2}$ (Quadrant I): $\sqrt {25-x^2}=5~cos~θ$
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