Precalculus (6th Edition) Blitzer

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

Chapter 6 - Review Exercises - Page 798: 35


$(5, \dfrac{3 \pi}{2})$

Work Step by Step

Here, $r=\sqrt {x^2+y^2}=\sqrt{(0)^2+(-5)^2}= \sqrt {25}=5$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{-5}{0}]$ Thus, $\theta =\dfrac{3 \pi}{2}$ Hence, $(5, \dfrac{3 \pi}{2})$
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