Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.1 Angles and Their Measure - 6.1 Assess Your Understanding - Page 361: 26

Answer

$\approx73.68°$

Work Step by Step

There are $60$ seconds in $1$ minute. Hence $40"=\frac{40}{60}'=0.6667'$ Hence $73°40'40"=73°(40+0.6667)'$ $=73°40.6667'$ There are $60$ minutes in $1$ degree. Hence $40.6667'=\frac{40.6667}{60}°=0.6778°$ Hence $73°40'40"=(73+0.6778)°$ $\approx 73.68°$ (to two decimal places)
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