Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - 5.3 Trigonometric Functions Values and Angle Measures - 5.3 Exercises - Page 533: 111

Answer

1.027349

Work Step by Step

Converting $13^{\circ}15'$ to decimal degrees, we get $13^{\circ}15'=13^{\circ}+\frac{15}{60}^{\circ}=13^{\circ}+0.25^{\circ}=13.25^{\circ}$ $\sec 13^{\circ}15'=\sec 13.25^{\circ}$ Now, we use the reciprocal identity $\sec\theta=\frac{1}{\cos\theta}$ to simplify the expression. $\sec 13.25^{\circ}=\frac{1}{\cos 13.25^{\circ}}=1.027349$
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