Precalculus (6th Edition)

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

Chapter 8 - Application of Trigonometry - 8.2 The Law of Cosines - 8.2 Exercises - Page 766: 12

Answer

$30^0$

Work Step by Step

From the law of cosines $1^{2}=1^{2}+\left( \sqrt {3}\right) ^{2}-2\times 1\times \left( \sqrt {3}\right) \times \cos \theta \Rightarrow 2\sqrt {3}\cos \theta =3\Rightarrow \cos \theta =\dfrac {\sqrt {3}}{2}\Rightarrow \theta =\cos ^{-1}\left( \dfrac {\sqrt {3}}{2}\right) =30^0$
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