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

Published by Pearson

# Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 694: 49

#### Answer

$10\text{ and } 10\sqrt{3}\approx17.321$

#### Work Step by Step

Let $a$ be the the side opposite the $30^o$ of the given right triangle. By the properties of a $30-60-90-$degree triangle, this side is half the hypotenuse. Hence $a=10.$ Let $c$ be the hypotenuse and $a,b$ be the legs of the right triangle. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, where $c=20$ and $a=10,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 10^2+b^2=20^2 \\\\ 100+b^2=400 \\\\ b^2=400-100 \\\\ b^2=300 \\\\ b=\sqrt{300} \\\\ b=\sqrt{100\cdot3} \\\\ b=\sqrt{(10)^2\cdot3} \\\\ b=10\sqrt{3} .\end{array} Hence, the missing measurements are $10\text{ and } 10\sqrt{3}\approx17.321$ units.

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