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 504: 32


$(3x-5)^\circ=40^\circ$ $(6x-40)^\circ=50^\circ$

Work Step by Step

If two angles are complementary, the sum of the two angles is a right angle ($90^\circ$): $(3x-5)^\circ+(6x-40)^\circ=(9x-45)^\circ=90^\circ$ $9x-45=90$ $9x=135$ $x=15$ $(3x-5)^\circ=40^\circ$ $(6x-40)^\circ=50^\circ$
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