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: 31


$(4 \sqrt 2, 135^{\circ})$

Work Step by Step

Here, $r=\sqrt {x^2+y^2}=\sqrt{(-4)^2+(4)^2}=4 \sqrt 2$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{4}{-4}]=arctan {-1}$ Thus, $\theta =135^{\circ}$ Hence, $(4 \sqrt 2, 135^{\circ})$
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