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 - Page 487: 5

Answer

$\sin\theta=\dfrac{40}{41}$ $\cos\theta=\dfrac{9}{41}$ $\tan\theta=\dfrac{40}{9}$ $\csc\theta=\dfrac{41}{40}$ $\sec\theta=\dfrac{41}{9}$ $\cot\theta=\dfrac{9}{40}$

Work Step by Step

The triangle is shown in the attached image below. Let $c$ be the unknown cathetus in the triangle. Use the Pythagorean Theorem to find $c$: $c=\sqrt{41^{2}-40^{2}}=\sqrt{1681-1600}=\sqrt{81}=9$ $\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{40}{41}$ $\cos\theta=\dfrac{adjacent}{hypotenuse}=\dfrac{9}{41}$ $\tan\theta=\dfrac{opposite}{adjacent}=\dfrac{40}{9}$ $\csc\theta=\dfrac{hypotenuse}{opposite}=\dfrac{41}{40}$ $\sec\theta=\dfrac{hypotenuse}{adjacent}=\dfrac{41}{9}$ $\cot\theta=\dfrac{adjacent}{opposite}=\dfrac{9}{40}$
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