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: 29

Answer

$40^{\circ} 19^{'} 12^{"}$

Work Step by Step

RECALL: $1^{\circ} = 60^{'} \text{ and } 1^{'} = 60^{"}$ \begin{equation} \begin{split} 40.32^{\circ} & = 40^{\circ}+0.32^{\circ} \\ & = 40^{\circ}+0.32 \times(1^{\circ}) \\ & = 40^{\circ}+0.32 \times (60^{'}) \\ & = 40^{\circ}+19.2 ^{'} \\ & = 40^{\circ}+19^{'}+0.2{'} \\ & = 40^{\circ}+19^{'}+0.2 \times (1)^{'} \\ & = 40^{\circ}+19^{'}+0.2 \times (60)^{"} \\ & = 40^{\circ}+19^{'}+12^{"} \end{split} \end{equation} Thus, $40.32^{\circ} = 40^{\circ} 19^{'} 12^{"}$
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