Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 677: 38


$9\sqrt{3} \text{ and } 18$

Work Step by Step

Since the given acute angle of the right triangle is $60^o,$ then the other acute angle measures $30^o.$ The side opposite this angle is half the hypotenuse. Hence, the hypotenuse is $18.$ The side opposite the $60^o$ angle, $a,$ is $\dfrac{\sqrt{3}}{2}\cdot c,$ where $c$ is the hypotenuse. Hence, \begin{array}{l}\require{cancel} \dfrac{\sqrt{3}}{2}\cdot c=a \\\\ \dfrac{\sqrt{3}}{2}\cdot 18=a \\\\ 9\sqrt{3}=a .\end{array} Hence, the missing sides have measure $ 9\sqrt{3} \text{ and } 18 $ units.
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