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

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