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 591: 43

Answer

$(-3\sqrt 3, 3)$

Work Step by Step

The point $(x,y)$ in the rectangular coordinate system can be expressed as: $x=r \ \cos(\theta)$, $y=r \ \sin(\theta) ...(1)$ Here, we have $r=6$ and $ \theta= 150^{\circ}$ Plug these values in equation (1) to obtain: $x=(6) \cos(150^{\circ})=(6)(-\dfrac{\sqrt 3}{2})=-3\sqrt 3 \\ y=(6) \sin(150^{\circ})=(6)(\dfrac{1}{2})=3$ Therefore, the point has coordinates $(-3\sqrt 3, 3)$ in the rectangular coordinate system.
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