Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.1 Angles and Their Measures - 5.1 Assess Your Understanding - Page 386: 23

Answer

$40.17^{\circ}$

Work Step by Step

$1^{\circ} = 1 \text{ Degree} \hspace{20pt} 1^{'} = 1 \text{ Minute} \hspace{20pt} 1^{"} = 1 \text{ Second}$ $\because 1^{'} = \left(\dfrac{1}{60}\right)^{\circ} \text{ and } 1^{"} =\left(\dfrac{1}{3600}\right)^{\circ} $ Thus, $40^{\circ} 10^{'} 25{"} = 40+\left(10 \times \dfrac{1}{60}\right) + \left(25 \times \dfrac{1}{3600}\right) \approx \boxed{40.17^{\circ}}$
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