Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.1 Polar Cordinates - 8.1 Assess Your Understanding - Page 592: 63



Work Step by Step

Solve for $r$ using the formula $r=\sqrt {x^{2}+y^{2}}$ to obtain: $$r=\sqrt {(1.3)^{2}+(-2.1)^{2}}=2.47$$ Since $(1.3, -2.1)$ is in the fourth quadrant, then $$\theta= \tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}\left(\frac{-2.1}{1.3}\right)\approx-1.02$$ Thus, the polar coordinates for the point $(1.3, -2.1)$ can be given as $(2.47,-1.02)$.
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