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: 14


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Work Step by Step

$\tan \theta =\dfrac{1}{\dfrac{6}{11}}=\dfrac{11}{6}$ $\tan \theta=\frac{\sin \theta}{\cos \theta}\\\rightarrow \sin \theta=\dfrac{11}{6}\cos \theta$ From the Pythagorean identity, we obtain: $$\sin^2 \theta +\cos^2 \theta=1\\(\dfrac{11}{6}\cos \theta)^2+\cos^2 \theta=1\\\cos^2 \theta=1-\dfrac{157}{36}\\\cos \theta=\sqrt\frac{36}{157}\\\\\cos \theta=\frac{6\sqrt 157}{157}$$ $\sec \theta=\dfrac{1}{\dfrac{6\sqrt 157}{157}}=\dfrac{\sqrt 157}{6}$ $\sin \theta=\dfrac{11}{6}\times\dfrac{6\sqrt 157}{157}=\dfrac{11\sqrt 157}{157}$ $\csc \theta=\dfrac{1}{\dfrac{11\sqrt 157}{157}}=\dfrac{\sqrt 157}{11}$
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