Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 9 - Polar Coordinates; Vectors - 9.1 Polar Coordinates - 9.1 Assess Your Understanding - Page 572: 45



Work Step by Step

By definition: $x=r\cdot \cos(\theta)$ and $y=r\cdot \sin(\theta)$. Here $r=6 \text{ and } \theta=150^\circ$. Hence, after plugging these in: $x=6\cdot \cos(150^\circ)=6\cdot(-\frac{\sqrt3}{2})=-3\sqrt3$ $y=6\cdot \sin(150^\circ)=6\cdot(\frac{1}{2})=3$ The point has coordinates $(-3\sqrt3,3)$ in the rectangular coordinate system.
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