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: 30

Answer

$(10x-20)^o=130^o ,\\ (x+15)^o=30^o ,\\ \left( x+5 \right)^o=20^o$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Equate the sum of the angles to $180$ and solve for $x.$ Then substitute the value of $x$ in the following given angles: \begin{array}{l}\require{cancel} (10x-20)^o ,\\ (x+15)^o ,\\ \left( x+5 \right)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of the angles of a triangle is always $180^o,$ then equate the sum of the given angles to $180.$ That is, \begin{array}{l}\require{cancel} (10x-20)+(x+15)+\left( x+5 \right)^o =180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 10x+x+x=180+20-15-5 \\\\ 12x=180 \\\\ x=\dfrac{180}{12} \\\\ x=15 .\end{array} Substituting $x= 15 $ in the angle $ (10x-20)^o $ results to \begin{array}{l}\require{cancel} (10\cdot15-20)^o \\\\= (150-20)^o \\\\= 130^o .\end{array} Substituting $x= 15 $ in the angle $ (x+15)^o $ results to \begin{array}{l}\require{cancel} (15+15)^o \\\\= 30^o .\end{array} Substituting $x= 15 $ in the angle $ (x+5)^o $ results to \begin{array}{l}\require{cancel} (15+5)^o \\\\= 20^o .\end{array} Hence, the measures of the angles of the triangle are \begin{array}{l}\require{cancel} (10x-20)^o=130^o ,\\ (x+15)^o=30^o ,\\ \left( x+5 \right)^o=20^o .\end{array}
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