Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set: 63

Answer

$\dfrac{\sqrt{3}}{3}$

Work Step by Step

RECALL: (1)The reference angle of a given angle is equal to the smallest acute angle that the terminal side makes with the x-axis. (2) Based on the location of the terminal side of an angle $\theta$, the reference angle can be found using the formula: (i) Quadrant I: $\theta$ (ii) Quadrant II: $180^0-\theta$ (iii) Quadrant III: $\theta-180^o$ (iv) Quadrant IV: $360^o-\theta$ The given angle is in Quadrant III. Use the formula in (iii) above to find its reference angle: $=210^0-180^o \\=30^o$ Note that $210^o$ is i Quadrant III, where tangent is positive. Thus, $\tan{210^o} = \tan{30^o}$ $30^o$ is a special angle whose tangent value is $\dfrac{\sqrt{3}}{3}$. Thus, $\tan{210^o} = \dfrac{\sqrt{3}}{3}$
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