Precalculus (6th Edition)

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

Chapter 5 - Trigonometric Functions - 5.1 Angles - 5.1 Exercises - Page 503: 20

Answer

(a) $50^{\circ}10^{\prime}$ (b) $140^{\circ}10^{\prime}$

Work Step by Step

(a) We can find the compliment by subtracting the given angle from $90^{\circ}$ which is equal to $89^{\circ}60^{\prime}$ $\implies 89^{\circ}60^{\prime}-39^{\circ}50^{\prime}=50^{\circ}10^{\prime}$ (b) We can find the supplement by subtracting the given angle from $180^{\circ}$ which is equal to $179^{\circ}60^{\prime}$ $\implies 179^{\circ}60^{\prime}-39^{\circ}50^{\prime}=140^{\circ}10^{\prime}$
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