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 - Test: Chapter 10 - Page 695: 23


$5 \text{ }cm\text{ and } 5\sqrt{3}\text{ } cm$

Work Step by Step

Let $a$ be the leg opposite the $30^o$ angle, $b$ be the leg opposite the $60^o$ angle, and $c=10$ be the hypotenuse. Using the properties of a $30-60-90$ degree triangle, then $a=5$ (half the hypotenuse) and $b=5\sqrt{3}$ ($\sqrt{3}/2$ times the hypotenuse.) Hence, the legs measure $ 5 \text{ }cm\text{ and } 5\sqrt{3}\text{ } cm .$
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