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: 36


$\dfrac{14\sqrt{3}}{3} \text{ and } \dfrac{7\sqrt{3}}{3}$

Work Step by Step

Since the given acute angle of the right triangle is $30^o,$ then the other acute angle is $60^o.$ The side opposite the $60^o$ 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 c=7 \\\\ c=7\cdot\dfrac{2}{\sqrt{3}} \\\\ c=7\cdot\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\\\ c=\dfrac{14\sqrt{3}}{3} .\end{array} The side opposite the $30^o$ angle is half the hypotenuse. With the hypotenuse equal to $\dfrac{14\sqrt{3}}{3},$ then the side opposite the $30^o$ angle is $ \dfrac{7\sqrt{3}}{3} .$ Hence, the missing sides have measure $ \dfrac{14\sqrt{3}}{3} \text{ and } \dfrac{7\sqrt{3}}{3} $ units.
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