Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Cumulative Review - Page 334: 35

Answer

The smallest angle is $30^o.$ The biggest angle is $110^o.$ And the remaining angle is $40^o.$

Work Step by Step

Let $x$ be the smallest angle. Then the largest angle is $x+80,$ and the remaining angle is $x+10.$ Since the sum of all angles of a triangle equal to $180^o,$ then \begin{array}{l}\require{cancel} x+(x+80)+(x+10)=180 .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 3x+90=180 \\ 3x=180-90 \\ 3x=90 \\ x=\dfrac{90}{3} \\ x=30 .\end{array} Hence, the smallest angle, $x,$ is $30^o.$ The biggest angle, $x+80,$ is $110^o.$ And the remaining angle, $x+10,$ is $40^o.$
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