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: 5


$\sin \theta=\dfrac{8}{11}$ $\cos \theta=\dfrac{\sqrt{57}}{11}$ $\tan \theta=\dfrac{8 \sqrt{57}}{57}$ $\csc \theta=\dfrac{11}{8}$ $\sec \theta=\dfrac{11\sqrt{57}}{57}$ $\cot \theta=\dfrac{8 \sqrt{57}}{57}$

Work Step by Step

We will have to use the Pythagorean Theorem to find the last side. Let $a$ be the last side, then $a=\sqrt {11^2-8^2}=\sqrt {121-64}=\sqrt{57}$ $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{8}{11}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{\sqrt{57}}{11}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{8}{\sqrt{57}}=\dfrac{8 \sqrt{57}}{57}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{11}{8}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{11\sqrt{57}}{57}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{8 \sqrt{57}}{57}$
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