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: 35

Answer

$(5x-1)^o=64^o ,\\ (2x)^o=26^o$

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} (5x-1)^o ,\\ (2x)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of complementary angles is $90^o,$ then \begin{array}{l}\require{cancel} (5x-1)+(2x)=90 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 5x+2x=90+1 \\\\ 7x=91 \\\\ x=\dfrac{91}{7} \\\\ x=13 .\end{array} Substituting $x= 13 $ in the angle $ (5x-1)^o $ results to \begin{array}{l}\require{cancel} (5\cdot13-1)^o \\\\= (65-1)^o \\\\= 64^o .\end{array} Substituting $x= 13 $ in the angle $ (2x)^o $ results to \begin{array}{l}\require{cancel} (2\cdot13)^o \\\\= 26^o .\end{array} Hence, the measures of the complementary angles are \begin{array}{l}\require{cancel} (5x-1)^o=64^o ,\\ (2x)^o=26^o .\end{array}
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