Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - 6.2 - Right Triangle Trigonometry - 6.2 Exercises - Page 444: 73


True. It is because $csc~60°=\frac{1}{sin~60°}$

Work Step by Step

$sin~60°=\frac{\sqrt 3}{2}$ (See the table on page 436) $csc~60°=\frac{1}{sin~60°}=\frac{1}{\frac{\sqrt 3}{2}}=\frac{2}{\sqrt 3}$ So: $sin~60°csc~60°=\frac{\sqrt 3}{2}\frac{2}{\sqrt 3}=1$
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