Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 1 - Trigonometric Functions - Section 1.4 Using the Definitions of the Trigonometric Functions - 1.4 Exercises - Page 38: 69

Answer

$\tan\theta=-\frac{\sqrt 3}{3}$

Work Step by Step

1.We get this information from the sine value $y=2$ $r=2$ 2. We can use y and r to find x (we need to remember that sign of x depends on quadrant, since it ends in quadrant II, x must be negative) $r=\sqrt {y^{2}+x^{2}}$ $y=-\sqrt {r^{2}-x^{2}}=-\sqrt {2^{2}-1^{2}}=-\sqrt {3}$ 4. Then insert values, $\tan\theta=\frac{-\sqrt 3}{3}$
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