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

Published by Pearson

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set: 33

Answer

$5\sqrt{3} \text{ and } 10\sqrt{3}$

Work Step by Step

Since the given acute angle of the right triangle is $60^o,$ then the side opposite this angle, $a,$ measures $\dfrac{\sqrt{3}}{2}\cdot c,$ where $c$ is the hypotenuse. Hence, the hypotenuse is \begin{array}{l}\require{cancel} \dfrac{\sqrt{3}}{2}\cdot c=a \\\\ c=15\cdot\dfrac{2}{\sqrt{3}} \\\\ c=15\cdot\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\\\ c=15\cdot\dfrac{2\sqrt{3}}{3} \\\\ c=10\sqrt{3} .\end{array} 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 $a=15$ and $c=10\sqrt{3} ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 15^2+b^2=(10\sqrt{3})^2 \\\\ \\\\ 225+b^2=100(3) \\\\ 225+b^2=300 \\\\ b^2=300-225 \\\\ b^2=75 \\\\ b=\sqrt{75} \\\\ b=\sqrt{25\cdot3} \\\\ b=\sqrt{(5)^2\cdot3} \\\\ b=5\sqrt{3} .\end{array} Hence, the missing sides have measure $5\sqrt{3} \text{ and } 10\sqrt{3}$ units.

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