Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.2 - Calculators and Trigonometric Functions of an Acute Angle - 2.2 Problem Set - Page 73: 89

Answer

$\sin\theta=-\frac{2\sqrt{13}}{13}$ $\cos\theta=\frac{3\sqrt{13}}{13}$ $\tan\theta = -\frac{2}{3}$

Work Step by Step

By definition $\sin\theta=\dfrac{y}{R}$ $\cos\theta=\dfrac{x}{R}$ $\tan\theta=\dfrac{y}{x}$ where $R^2=x^2+y^2.$ Our point has the coordinates $x=3$ and $y=-2$. then $R=\sqrt{3^2+(-2)^2}=\sqrt{13}$. So $\sin\theta=\dfrac{-2}{\sqrt{13}}=-\dfrac{2\sqrt{13}}{13}$ $\cos\theta=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}$ $\tan\theta = -\dfrac{2}{3}$
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