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: 37


$14\sqrt{3} \text{ and } 28$

Work Step by Step

Since the given acute angle of the right triangle is $30^o,$ then the side opposite this angle is half the hypotenuse. Hence, the hypotenuse is $28.$ The other acute angle measures $60^o.$ The side opposite this 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 28=a \\\\ 14\sqrt{3}=a .\end{array} Hence, the missing sides have measure $ 14\sqrt{3} \text{ and } 28 $ units.
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