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: 36



Work Step by Step

Here, $r=\sqrt {x^2+y^2}=\sqrt{(1)^2+(0)^2}= 1$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{0}{1}]$ Thus, $\theta =0$ Hence, $(1,0)$
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