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: 32


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

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 the $30^o$ angle measures $ 9 $ (half the given hypotenuse, $18.$) Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $ a=9 $ and $ c=18 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 9^2+b^2=18^2 \\\\ 81+b^2=324 \\\\ b^2=324-81 \\\\ b^2=243 \\\\ b=\sqrt{243} \\\\ b=\sqrt{81\cdot3} \\\\ b=\sqrt{(9)^2\cdot3} \\\\ b=9\sqrt{3} .\end{array} Hence, the missing sides have measure $ 9 \text{ and } 9\sqrt{3} $ units.
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