Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.1 Use Trigonometry with Right Triangles - 13.1 Exercises - Skill Practice - Page 856: 11


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Work Step by Step

$\cot \theta=\dfrac{1}{\dfrac{7}{3}}=\dfrac{3}{7}$ $\tan \theta=\dfrac{\sin \theta}{\cos \theta}\\\rightarrow\sin \theta=\tan \theta \times\cos \theta\\\rightarrow \sin \theta=\dfrac{7}{3}\cos \theta$ From the Pythagorean identity, we obtain: $$\sin^2 \theta +\cos^2 \theta=1\\(\dfrac{7}{3}\cos \theta)^2+\cos^2 \theta=1\\\dfrac{49}{9}\cos^2+\cos^2 \theta=1\\ \dfrac{58}{9}\cos^2 \theta=1\\\cos^2\theta=\frac{9}{58}\\\\\cos \theta=\sqrt \frac{9}{58}=\frac{3\sqrt 58}{58}$$ $\sin \theta=\dfrac{7}{3}\times\dfrac{3\sqrt 58}{58}=\dfrac{7\sqrt 58}{58}$ $\sec \theta=\dfrac{1}{\dfrac{3\sqrt 58}{58}}=\dfrac{\sqrt 58}{3}$ $\csc \theta=\dfrac{1}{\dfrac{7\sqrt 58}{58}}=\dfrac{58}{7\sqrt 58}=\dfrac{\sqrt 58}{7}$
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