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: 4


$\sin \theta=\dfrac{3 \sqrt{73}}{73}$ $\cos \theta=\dfrac{8\sqrt{73}}{73}$ $\tan \theta=\dfrac{3}{8}$ $\csc \theta=\dfrac{\sqrt{73}}{3}$ $\sec \theta=\dfrac{\sqrt{73}}{8}$ $\cot \theta=\dfrac{8}{3}$

Work Step by Step

Use the Pythagorean Theorem to find the Hypotenuse. $Hypotenuse=\sqrt {3^2+8^2}=\sqrt {9+84}=\sqrt{73}$ $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{3}{\sqrt{73}}=\dfrac{3 \sqrt{73}}{73}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{8}{\sqrt{73}}=\dfrac{8\sqrt{73}}{73}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{3}{8}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{\sqrt{73}}{3}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{\sqrt{73}}{8}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{8}{3}$
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