Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.2 - Trigonometry of Right Triangles - 6.2 Exercises: 6

Answer

$\sin\theta=\dfrac{15}{17}$ $\cos\theta=\dfrac{8}{17}$ $\tan\theta=\dfrac{15}{8}$ $\csc\theta=\dfrac{17}{15}$ $\sec\theta=\dfrac{17}{8}$ $\cot\theta=\dfrac{8}{15}$

Work Step by Step

The triangle is shown in the attached image below. Let $h$ be the hypotenuse of the triangle. Use the Pythagorean Theorem to find it: $h=\sqrt{8^{2}+15^{2}}=\sqrt{64+225}=\sqrt{289}=17$ $\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{15}{17}$ $\cos\theta=\dfrac{adjacent}{hypotenuse}=\dfrac{8}{17}$ $\tan\theta=\dfrac{opposite}{adjacent}=\dfrac{15}{8}$ $\csc\theta=\dfrac{hypotenuse}{opposite}=\dfrac{17}{15}$ $\sec\theta=\dfrac{hypotenuse}{adjacent}=\dfrac{17}{8}$ $\cot\theta=\dfrac{adjacent}{opposite}=\dfrac{8}{15}$
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