Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - Chapter 8 Test Prep - Review Exercises - Page 842: 96

Answer

$$r = 2\csc \theta $$

Work Step by Step

$$\eqalign{ & y = 2 \cr & {\text{Where }}r\sin \theta = y,{\text{ then}} \cr & r\sin \theta = 2 \cr & r = \frac{2}{{\sin \theta }} \cr & {\text{Simplify}} \cr & r = 2\csc \theta \cr} $$
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