Precalculus (6th Edition)

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

Chapter 8 - Application of Trigonometry - 8.4 Algebraically Defined Vectors and the Dot Product - 8.4 Exercises - Page 789: 2

Answer

$30^{0}$

Work Step by Step

$\tan \theta =\dfrac {1}{\sqrt {3}}\Rightarrow \theta =\tan ^{-1}\theta \dfrac {1}{\sqrt {3}}=30^{0}$
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