Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises: 12

Answer

$\sin \theta =\dfrac {\sqrt {11}}{6}$

Work Step by Step

$\sin \theta =\pm \sqrt {1-\cos ^{2}\theta }=\pm \sqrt {1-\left( \dfrac {5}{6}\right) ^{2}}=\pm \dfrac {\sqrt {11}}{6}$ In quadrant 1 sine is positive so $\sin \theta =\dfrac {\sqrt {11}}{6}$
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