Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.4 - Further Applications of Linear Equations - 1.4 Exercises: 36

Answer

$(3x-9)^o=24^o ,\\ (6x)^o=66^o$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Equate the given complementary angles to $90$, and use the properties of equality to isolate $x.$ Then substitute the value of $x$ in the following given complementary angles: \begin{array}{l}\require{cancel} (3x-9)^o ,\\ (6x)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of complementary angles is $90^o,$ then \begin{array}{l}\require{cancel} (3x-9)+(6x)=90 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x+6x=90+9 \\\\ 9x=99 \\\\ x=\dfrac{99}{9} \\\\ x=11 .\end{array} Substituting $x= 11 $ in the angle $ (3x-9)^o $ results to \begin{array}{l}\require{cancel} (3\cdot11-9)^o \\\\= (33-9)^o \\\\= 24^o .\end{array} Substituting $x= 11 $ in the angle $ (6x)^o $ results to \begin{array}{l}\require{cancel} (6\cdot11)^o \\\\= 66^o .\end{array} Hence, the measures of the complementary angles are \begin{array}{l}\require{cancel} (3x-9)^o=24^o ,\\ (6x)^o=66^o .\end{array}
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