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: 6


$\sin \theta=\dfrac{7}{15}$ $\cos \theta=\dfrac{4 \sqrt{11}}{15}$ $\tan \theta=\dfrac{7\sqrt{11}}{44}$ $\csc \theta=\dfrac{15}{7}$ $\sec \theta=\dfrac{15\sqrt{11}}{44}$ $\cot \theta=\dfrac{4\sqrt{11}}{7}$

Work Step by Step

We will have to use the Pythagorean Theorem to find the last side. Let $a$ be the last side, then $a=\sqrt {15^2-7^2}=\sqrt {225-49}=4 \sqrt{11}$ $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{7}{15}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{4 \sqrt{11}}{15}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{7}{4\sqrt{11}}=\dfrac{7\sqrt{11}}{44}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{15}{7}$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{15\sqrt{11}}{44}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{4\sqrt{11}}{7}$
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